Show that the points (a, a),(-a, - a) and (-√3a, √3a) are the vertices of an equilateral triangle. Also, find its area.
Answers
Let they be
A(a,a)B(-a,-a)C(-√3a,√3a)Find distances
\(\\ \rm\Rrightarrow AB=\sqrt{(a+a)^2+(a+a)^2}=\sqrt{4a^2+4a^2}=\sqrt{8a^2}=2\sqrt{2}a\)
\(\\ \rm\Rrightarrow BC=\sqrt{(-√3a+a)^2+(√3a+a)^2}\)
(a+b)^2+(a-b)^2=2a²+2b²\(\\ \rm\Rrightarrow BC=\sqrt{2(\sqrt{3}a)^2+2a^2}=\sqrt{6a^2+2a^2}=\sqrt{8a^2}=2\sqrt{2}a\)
\(\\ \rm\Rrightarrow AC=\sqrt{(a+\sqrt{3}a)^2+(a-\sqrt{3}a)^2}=2\sqrt{2}a\)
As
AB=AC=BCThe traingle is equilateral
Area
\(\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}a^2\)
\(\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}(2\sqrt{2}a)^2\)
\(\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}(8a^2)\)
\(\\ \rm\Rrightarrow 2\sqrt{3}a^2\)
\( {\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}\)
Let, the given points be
\(\bullet \: \tt A (a, a)\)
\(\bullet \: \tt B (-a, -a)\)
\(\bullet \: \tt C ( - \sqrt{3}a , \sqrt{3}a)\)
Now,
Finding the distances of AB, BC and CD using distance formula.
\(\star \: \bold{\underline \red{\sf{ Finding \: AB }}}\)
Using,
\( \tt Distance \: Formula = \sqrt{{(x{\small_{2}} -x{\small_{1}} )}^{2} +{(y{\small_{2}} -y{\small_{1}} )}^{2}}\)
where
\(\sf x{\small_{2}} = - a\)
\(\sf x{\small_{1}} = a\)
\(\sf y{\small_{2}} = - a\)
\(\sf y{\small_{1}} = a\)
Putting the values,
\(\longrightarrow \sf AB = \sqrt{( { - a - a)}^{2} + {( - a - a)}^{2} }\)
\(\longrightarrow \sf AB = \sqrt{( { - 2 a)}^{2} + {( -2 a)}^{2} }\)
\(\longrightarrow \sf AB = \sqrt{4{ a}^{2} + 4{ a}^{2} }\)
\(\longrightarrow \sf AB = \sqrt{8{ a}^{2} }\)
\(\longrightarrow \sf AB = \sqrt{2 \times 2 \times 2 \: { a}^{2} }\)
\(\longrightarrow \sf AB = \red{2 \sqrt{ 2 }a}\)
\( \star \: \bold{\underline \green{\sf{ Finding \: BC }}}\)
Using,
\( \tt Distance \: Formula = \sqrt{{(x{\small_{2}} -x{\small_{1}} )}^{2} +{(y{\small_{2}} -y{\small_{1}} )}^{2}}\)
where
\(\sf x{\small_{2}} = - \sqrt{3}a \)
\(\sf x{\small_{1}} = - a\)
\(\sf y{\small_{2}} = \sqrt{3}a \)
\(\sf y{\small_{1}} = - a\)
Putting the values,
\( \longrightarrow \sf BC = \sqrt{ \bigg[ { - \sqrt{3}a - (- a) \bigg]}^{2} + {\bigg[ \sqrt{3}a - (- a)\bigg]}^{2} }\)
\( \longrightarrow \sf BC = \sqrt{ { ( - \sqrt{3}a + a) }^{2} + {( \sqrt{3} a + a)}^{2} }\)
Using (a + b)² = a² + 2(a)(b) + b².
\(\longrightarrow \sf BC = \sqrt{ { (- \sqrt{3}a) }^{2} + 2( - \sqrt{3} a)(a) + {(a)}^{2} + {( \sqrt{3} a + a)}^{2} }\)
\(\longrightarrow \sf BC = \sqrt{ 3 {a}^{2} - 2 \sqrt{3} {a}^{2} + {a}^{2} + {( \sqrt{3} a + a)}^{2} }\)
Again, using (a + b)² = a² + 2(a)(b) + b².
\(\longrightarrow \sf BC = \sqrt{ 3 {a}^{2} - 2 \sqrt{3} {a}^{2} + {a}^{2} + { ( \sqrt{3}a) }^{2} + 2( \sqrt{3} a)(a) + {(a)}^{2} }\)
\(\longrightarrow \sf BC = \sqrt{ 3 {a}^{2} \:\cancel{ - 2 \sqrt{3} {a}^{2}} + {a}^{2} + 3 {a}^{2} \:\cancel{ + 2 \sqrt{3} {a}^{2}} + {a}^{2} }\)
\( \longrightarrow \sf BC = \sqrt{ 3 {a}^{2} + {a}^{2} + 3 {a}^{2} + {a}^{2} }\)
\( \longrightarrow \sf BC = \sqrt{ 8 {a}^{2} }\)
\(\longrightarrow \sf BC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }\)
\( \longrightarrow \sf BC = \green{2 \sqrt{ 2 }a}\)
\( \star \: \bold{\underline \orange{\sf{ Finding \: AC }}}\)
Using,
\( \tt Distance \: Formula = \sqrt{{(x{\small_{2}} -x{\small_{1}} )}^{2} +{(y{\small_{2}} -y{\small_{1}} )}^{2}}\)
where
\(\sf x{\small_{2}} = - \sqrt{3}a \)
\(\sf x{\small_{1}} = a\)
\(\sf y{\small_{2}} = \sqrt{3}a \)
\(\sf y{\small_{1}} = a\)
Putting the values,
\( \longrightarrow \sf AC = \sqrt{ { ( - \sqrt{3}a - a)}^{2} + {(\sqrt{3}a - a)}^{2} }\)
Using (a - b)² = a² - 2(a)(b) + b².
\( \longrightarrow \sf AC = \sqrt{ {( - \sqrt{3} a)}^{2} - 2( - \sqrt{3} a)(a) + {(a)}^{2} + {(\sqrt{3}a - a)}^{2} }\)
\( \longrightarrow \sf AC = \sqrt{ 3 {a}^{2} + 2 \sqrt{3} {a}^{2} + {a}^{2} + {(\sqrt{3}a - a)}^{2} }\)
Again, using (a - b)² = a² - 2(a)(b) + b².
\(\longrightarrow \sf AC = \sqrt{ 3 {a}^{2} + 2 \sqrt{3} {a}^{2} + {a}^{2} + {( \sqrt{3} a)}^{2} - 2( \sqrt{3}a)(a) + {(a)}^{2} }\)
\(\longrightarrow \sf AC = \sqrt{ 3 {a}^{2} \: \cancel{ + 2 \sqrt{3} {a}^{2} } + {a}^{2} + 3 {a}^{2} \: \cancel{ - 2 \sqrt{3} {a}^{2} } + {a}^{2} }\)
\(\longrightarrow \sf AC = \sqrt{ 3 {a}^{2} + {a}^{2} + 3 {a}^{2} + {a}^{2} }\)
\(\longrightarrow \sf AC = \sqrt{ 8 {a}^{2} }\)
\( \longrightarrow \sf AC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }\)
\( \longrightarrow \sf AC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }\)
\(\longrightarrow \sf AC = \orange{{2\sqrt{2 }}a}\)
Since,
AB = BC = CA = \( \sf {{2\sqrt{2 }}a} \)
Therefore, the \(\triangle \) ABC formed by the given points is an equilateral triangle.
For,
Finding the area of \(\triangle \) ABC
Using,
\(\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4} \: {(side)}^{2} \)
Putting,
• side = \( \sf {{2\sqrt{2 }}a} \)
\(\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4} \: {(2 \sqrt{2}a )}^{2} \)
\(\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4 } \times 2 \times 2 \times 2 \times {a}^{2} \)
\(\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{ \cancel{4}} \times \cancel{8 }{a}^{2} \)
\(\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \purple{2 \sqrt{3} {a}^{2} \: sq. \: units}\)
\(\therefore \sf Area = 2 \sqrt{3} {a}^{2} \: sq. \: units\)
\( {\underline{\rule{310pt}{2pt}}} \)
Related Questions
Part A Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft x 15.0 ft x 9.10ft . Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 16.4 g Submit Previous Answers Correct Part B Consider the formation of HCN by the reaction of NaCN (sodium cyanide) with an acid such as H2SO4 (sulfuric acid): 2NaCN(s) + H2SO4 (aq) +Na2SO4 (aq) + 2HCN(g) What mass of NaCN gives the lethal dose in the room? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 29.8 g Submit Previous Answers Correct Correct answer is shown. Your answer 29.798 g was either rounded differently or used a different number of significant figures than required for this part. Part C HCN forms when synthetic fibers containing Orlon® or Acrilan® burn. Acrilan® has an empirical formula of CH, CHCN, so HCN is 50.9% of the formula by mass. A rug in the laboratory measures 12.0x 12.0 ft and contains 30.0 oz of Acrilan® fibers per square yard of carpet. If the rug burns, what mass of HCN will be generated in the room? Assume that the yield of HCN from the fibers is 20.0% and that the carpet is 40.0 % consumed. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 0 uÅ ? 1088.624 g Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Your answer implies that Acrilan® is 100% HCN. Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately 300. mg HCN per kilogram of air when inhaled. The density of air at 26 °C is 0.00118 g/cm'. 3 .
Answers
Part A: To calculate the amount of HCN that gives the lethal dose in a small laboratory room, we need to determine the volume of the room first. The volume of the room can be calculated by multiplying the length, width, and height of the room.
Given:
Length = 12.0 ft
Width = 15.0 ft
Height = 9.10 ft
Volume = Length × Width × Height
Plugging in the values, we get:
Volume = 12.0 ft × 15.0 ft × 9.10 ft
Now, we can convert the volume from cubic feet to liters using the conversion factor: 1 ft^3 = 28.32 L.
Volume = (12.0 ft × 15.0 ft × 9.10 ft) × (28.32 L/1 ft^3)
Next, we need to calculate the lethal dose of HCN per kilogram of air. The lethal dose is approximately 300 mg HCN per kilogram of air.
Now, we can convert the volume from liters to kilograms using the density of air at 26 °C, which is 0.00118 g/cm^3.
Mass of air = Volume × Density of air
Mass of air = Volume × (0.00118 g/cm^3 × 1000 kg/g)
Finally, we can calculate the amount of HCN that gives the lethal dose by multiplying the mass of air by the lethal dose per kilogram of air.
Amount of HCN = Mass of air × Lethal dose per kilogram of air
Expressing the answer to three significant figures, the amount of HCN that gives the lethal dose in the room is X grams.
Part B: To calculate the mass of NaCN that gives the lethal dose in the room, we need to use the balanced chemical equation for the reaction of NaCN with H2SO4.
The equation is:
2NaCN(s) + H2SO4(aq) → Na2SO4(aq) + 2HCN(g)
From the equation, we can see that 2 moles of NaCN react to form 2 moles of HCN. Therefore, the molar ratio between NaCN and HCN is 2:2.
Now, we can calculate the molar mass of NaCN, which is the sum of the atomic masses of sodium (Na), carbon (C), and nitrogen (N).
Molar mass of NaCN = (Atomic mass of Na) + (Atomic mass of C) + (Atomic mass of N)
Next, we need to calculate the number of moles of HCN needed to give the lethal dose in the room. We can use the molar ratio between NaCN and HCN to determine this.
Number of moles of HCN = Number of moles of NaCN × (2 moles of HCN / 2 moles of NaCN)
Finally, we can calculate the mass of NaCN using the molar mass and the number of moles of NaCN.
Mass of NaCN = Number of moles of NaCN × Molar mass of NaCN
Expressing the answer to three significant figures, the mass of NaCN that gives the lethal dose in the room is X grams.
Part C: To calculate the mass of HCN generated in the room when the rug burns, we need to consider the mass of Acrilan® fibers and the yield of HCN from the fibers.
Given:
Rug area = 12.0 ft × 12.0 ft
Mass of Acrilan® fibers per square yard of carpet = 30.0 oz
Yield of HCN from the fibers = 20.0%
Carpet consumed = 40.0%
First, we need to calculate the mass of Acrilan® fibers in the rug. We can use the area of the rug and the mass of fibers per square yard of carpet to determine this.
Mass of Acrilan® fibers in the rug = Rug area × (Mass of fibers per square yard of carpet / Area of one square yard)
Next, we can calculate the mass of HCN generated from the Acrilan® fibers by multiplying the mass of fibers by the percentage of HCN in the formula (50.9%).
Mass of HCN generated = Mass of Acrilan® fibers × Percentage of HCN in the formula
Now, we need to consider the yield of HCN and the carpet consumed. We can calculate the actual mass of HCN generated in the room by multiplying the mass of HCN generated by the yield and the percentage of carpet consumed.
Actual mass of HCN generated = Mass of HCN generated × (Yield of HCN / 100) × (Carpet consumed / 100)
Expressing the answer to three significant figures, the mass of HCN generated in the room when the rug burns is X grams.
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Determine the change to the base of a triangle if the height is tripled and the area is 12 times larger than the original.
I'll give brainliest to first answer. Help please
Answers
Answer:
If the height is tripled and the area is 12 times larger, the base would be 8 times larger.
Step-by-step explanation:
Area of Triangle = 1/2(base × height)
If the area is 1 foot and the height is 1 foot, that means the base is 2 feet, since 1 = 1/2(2 × 1)
Now, if the height is tripled and the area is 12 times larger, then the base would have to be 8 feet, since 12 = 1/2(8 × 3)
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes. nothing (Simplify your answer. Round to three decimal places as needed.)
Answers
The probability that a randomly selected passenger has a waiting time greater than 1.25 minutes is 0.792.
The waiting times are uniformly distributed between 0 and 6 minutes. To find the probability of a waiting time greater than 1.25 minutes, we need to consider the remaining time in the interval, which is 6 - 1.25 = 4.75 minutes.
Since the total waiting time is uniformly distributed over 6 minutes, the probability of a waiting time greater than 1.25 minutes is the ratio of the remaining time to the total time. So, the probability is:
P(waiting time > 1.25) = (4.75 minutes) / (6 minutes) = 0.7917
Rounded to three decimal places, the probability is 0.792.
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Please help with this
Answers
1) (a) The transformations that occur from the parent function are horizontal translation of 2 units to the left and vertical translation of 4 units to the down.
(b) (-2, -4)
(c) Graph is given below.
1) Given a function,
g(x) = (x + 2)² - 4
(a) Given a parent function p(x) = x².
We can write g(x) as,
g(x) = p(x + 2) - 4
So the transformation is horizontal translation of 2 units to the left and vertical translation of 4 units to the down.
(b) Vertex formula of a parabola is,
y = a (x - h)² + k, where (h, k) is the vertex.
Comparing the given function with vertex form,
Vertex of the parabola = (-2, -4)
(c) Graph of g(x) will be a parabola with vertex at (-2, -4).
It is given below.
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Select the correct number from each drop-down menu to show the values of and y.
Answers
The angle x of the triangle is 68 degrees and the side y is 7.5 units.
How to find the angles of a triangle?The triangle ABC is a right angle triangle because it has one angle as 90 degrees. The sum of the angles in the triangle is 180 degrees.
The angle ECD is congruent to angle ACB(vertically opposite angles).
Therefore,
∠ECD ≅ ∠ACB
Hence,
x = 180 - 90 - 22
x = 68 degrees.
using trigonometric ratios, let's find y
cos 22 = adjacent / hypotenuse
sin 22 = 2y - 1 / 15
cross multiply
15 cos 22 = 2y - 1
2y - 1 = 13.9077578185
2y = 13.9077578185 + 1
2y = 14.9077578185
divide both sides by 2
y = 14.9077578185 / 2
y = 7.45387890925
Therefore,
y = 7.5 units
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A backpack that normally sells for $39 is on sale for $25. Find the
percent of change.
Answers
Answer: To find the discount, simply multiply the original selling price by the %discount:
ie: 39 x 33/100= $12.87
So, the discount is $12.87.
Step-by-step explanation: To find the sale price, simply minus the discount from the original selling price:
ie: 39- 12. 87= 26.13
So, the sale price is $26.13
What is the value of x in the figure below? In this diagram, AABD~ ACAD.
12
X
27
Answers
Step-by-step explanation:
Based on the information given in the diagram, we know that AABD and ACAD are similar triangles, which means their corresponding sides are proportional.
Let's denote the length of AD as x. According to the similarity of the triangles, we can set up the following proportion:
AB/AC = AD/AD
Since AD is common to both triangles and has a length of x, the proportion simplifies to:
AB/AC = 1
We are given that AB = 12, so we can substitute that value into the proportion:
12/AC = 1
To solve for AC, we can cross-multiply:
12 = AC
Therefore, the value of x in the figure is 12.
Let T be total amount (in dollars) saved using both plans combined. Write an equation relating T to N. Simplify your answer as much as possible.Hong and his wife are each starting a saving plan. Hong will initially set aside $250 and then add \$155 every month to the savings. The amount A (in dollars) saved this way is given by the function A=155N-25O, where N is the number of months he has been saving His wife will not set an initial amount aside but will add $385 to the savings every month. The amount B (in dollars) saved using this plan is given by the function B = 385N
Answers
Answer:
T = 540N + 250
Explanation:
The amount saved by Hong after N months is
A = 155N + 250
And the amount saved by his wife after N months is
B = 385N
Then, the total amount saved is the sum of these expressions, so
T = A + B
T = (155N + 250) + (385N)
Simplifying, we get
T = 155N + 250 + 385N
T = 540N + 250
Therefore, the answer is
T = 540N + 250
The area of a rectangle is given as x2+5x+6. which expression represents either the length or widith of the rectangle. these are the answers, it has to be one of these (x-3) (x+1) (x+3) (x+6)
Answers
The expression which represents either the length or width is (x+3). The correct answer is option (c).
Given that the area of a rectangle which is given as x²+5x+6.
We have to find which expression represents either the length or width.
We are given an expression for the area. But we also know that the area of a rectangle is length times width. So the expression we were given, x²+5x+6 must be equal to the length of the rectangle times its width.
So, will find the factoring the expression we can see expressions for the length and the width.
Factoring x²+5x+6 we get (x+2)(x+3).
One of these factors is the width and the other is the length.
Hence, the expression represents either the length or width of the rectangle when the area of a rectangle is given as x²+5x+6 is (x+3).
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5^(9x)=25^(4x+2) SHOW STEPSSSS
Answers
Answer:
x = 4
Step-by-step explanation:
\(5^{9x} =25^{4x+2}\)
First you need to get a common base as your number.
\(5^{9x} =5^{2(4x+2)}\)
The second number changed because 5 squared is 25. Changing it to this form gives us the same base number of 5. Now we use the distributive property in the exponent.
\(5^{9x} =5^{8x+4}\)
Now we can use this statement:
\(A^{m} =A^{n}\) if \(m=n\)
Using this we solve:
9x = 8x+4
-8x -8x
x = 4
Hope it helps!
12
Find the lowest common multiple (LCM) of 28, 42 and 63
Show your working clearly.
Answers
Answer:
Least Common Multiple (LCM) of 28,42,63 is 252 ∴ So the LCM of the given numbers is 2 x 3 x 7 x 2 x 1 x 3 = 252
Step-by-step explanation:
Answer:
252 is the answer
Step-by-step explanation:
find the multiples of all of them ( and make sure it is the least. )
28:
28, 56, 84, 112, 140, 168, 196, 224, 252, 280, 308
42:
42, 84, 126, 168, 210, 252, 294, 336
63:
63, 126, 189, 252, 315, 378
bolded + undurlined is the answer
you see that 252 is the answer
252 is the answer
For each of the following functions, what value must the constant c have in order for it to be a probability mass function? (a) p(x) = c/x where x = - 1,2,3 (b) p(x) = cx^2 where x = 1,2,3,4
Answers
The constant c must be 1/30 for p(x) = cx² to be a probability mass function
(a) For p(x) = c/x! to be a PMF, c must be 0.6.
(b) For p(x) = cx² to be a PMF, c must be 1/30.
To be a probability mass function (PMF), a function must satisfy two conditions:
The sum of all probabilities must equal 1.
The probabilities must be non-negative for all possible values of the random variable.
Let's analyze each function separately:
(a) For the function p(x) = c/x!, where x = -1, 2, 3:
To find the value of c, we need to ensure that the sum of probabilities equals 1. Let's calculate it:
p(-1) + p(2) + p(3) = c/(-1)! + c/2! + c/3!
To simplify the expression, we'll calculate the factorials:
p(-1) + p(2) + p(3) = c/1 + c/2 + c/6
Now, we need to set this sum equal to 1 and solve for c:
c/1 + c/2 + c/6 = 1
To find the common denominator, multiply each term by 6:
6c + 3c + c = 6
Combine like terms:
10c = 6
Divide both sides by 10:
c = 0.6
So, the constant c must be 0.6 for p(x) = c/x! to be a probability mass function.
(b) For the function p(x) = cx² where x = 1, 2, 3, 4:
Again, we need to ensure that the sum of probabilities equals 1. Let's calculate it:
p(1) + p(2) + p(3) + p(4) = c(1²) + c(2²) + c(3²) + c(4²)
Simplifying the expression:
p(1) + p(2) + p(3) + p(4) = c + 4c + 9c + 16c
Combining like terms:
p(1) + p(2) + p(3) + p(4) = 30c
To find the value of c, we need to set this sum equal to 1 and solve for c:
30c = 1
Divide both sides by 30:
c = 1/30
Therefore, the constant c must be 1/30 for p(x) = cx² to be a probability mass function
(a) For p(x) = c/x! to be a PMF, c must be 0.6.
(b) For p(x) = cx² to be a PMF, c must be 1/30.
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can someone please help me with question 6
Answers
Answer:
5323
Step-by-step explanation:
453
Please help, having trouble figuring out this word problem. Thanks if you do :)
Answers
Answer:
-21. -19.-17
Step-by-step explanation:
Let x be the smallest
x+2 next smallest
x+4 be the largest
4*(x) = 18+6(x+4)
Distribute
4x = 18+6x+24
Combine like terms
4x = 42+6x
Subtract 6x from each side
-2x = 42
Divide by -2
-2x/-2 = 42/-2
x = -21
x+2 = -19
x+4 = -17
Answer:
- 21, - 19, - 17
Step-by-step explanation:
Let y be the smallest no.
& y + 2 next smallest no.
& y + 4 be the largest no.
= 4 * (y) = 18 + 6 * (y + 4)
4y = 18 + 6y + 24
4y - 6y = 42
- 2y = 42
= - 21
The other 2 numbers are:
y + 2
= - 19
y + 4
= - 17
Hope this helps!
Prediction of the value of the dependent variable outside the experimental region is called _____. a. extrapolation b. averaging c. interpolation d. forecasting
Answers
The prediction of the value of the dependent variable outside the experimental region is called
extrapolation. So, the option(a) is right one.
A dependent variable is defined as the variable which is tested and measured in a scientific experiment. It is always depends on other variables. That's why it is called dependent variable and other variable is independent variable. Because it is a variable so it's value always change according to situation. So, there are two processes for predicting the values of dependent variable. These are defined as below :
The process of predicting inside of the observations of x values observed in the data is called interpolation. The process of predicting outside of the observations x values observed in the data is called extrapolation.Hence, the prediction of the value of the dependent variable outside the experimental region is known as extrapolation.
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Determine whether each scatter plot represents a linear relationship, a non-linear relationship, or
no relationship
Answers
What is the remainder when when f(x)=x3+3x2−10x−14 is divided by (x-4) ?
Answers
Answer:
58
Step-by-step explanation:
With Horner's Method.
^^
The remainder when f(x) is divided by (x - 4) is 58.
What is Remainder?The value remaining after division is known as the Remainder. After division, we are left with a value if a number (dividend) cannot be divided entirely by another number (divisor). The remainder is the name for this amount.
As per the given data:
f(x) = x³ + 3x² - 10x - 14
To find the remainder when f(x) is divided by (x - 4).
Using the remainder theorem:
When f(x) is divided by (x - a) then the remainder is given by f(a).
Remainder when f(x) is divided by (x - 4):
f(4) = (4)³ + 3(4)² - 10(4) - 14
f(4) = 58
Hence, the remainder when f(x) is divided by (x - 4) is 58.
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By what percent will a fraction CHANGE(not increase!!) if it’s numerator is decreased by 60% and it’s denominator is decreased by 20%?
I am aware that it is a decrease, but not sure of how much it is a decrease by.
HELP ASAP!!
Answers
if i was i your class or in bigger class i will help you
pls help ASAP !! 30 PTS
Answers
Answer: C = 100m + 25
Step-by-step explanation: Since c and m were the variable here and $100. Initial fee is m in this case.
The equation is C = 100m + 25
How long would it take R20000 invested today at a simple interest rate of 9% p.a. to reach an investment goal of R30000.
A Approximately 5.6 years
B Approximately 6.1 years
C Approximately 4.7 years
D Approximately 5.1 years
Answers
\(~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 30000\\ P=\textit{original amount deposited}\dotfill & \$20000\\ r=rate\to 9\%\to \frac{9}{100}\dotfill &0.09\\ t=years \end{cases} \\\\\\ 30000 = 20000[1+(0.09)(t)] \implies \cfrac{30000}{20000}=1+0.09t\implies \cfrac{3}{2}=1+0.09t \\\\\\ \cfrac{3}{2}-1=0.09t\implies \cfrac{1}{2}=0.09t\implies \cfrac{1}{2(0.09)}=t\implies 5.6\approx t\)
scatterplot strength and form: which one of the four scatterplots below shows a strong negative linear association between two variables?
Answers
A strong negative linear association between two variables is scatter plot 4.
What is standard normal random?
We shall read probabilities out rather than directly working with Z's density function to compute probabilities for Z. The elements in the tables are probabilities of the kind P(Zz), and they are tables of cumulative probabilities. Following are a number of examples that will help to clarify how to use the tables.The standard normal random.
Very dispersed and lacking in association is scatter plot 1.
Positive correlation may be seen in scatter plots 2 and 3.
Scatter figure #4 clearly illustrates a strong inverse relationship between the variables.
Hence the Correct answer is scatter plot 4.
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EspanolA lamp has the shape of a parabola when viewed from the side. The light source (which is at the focus) is 3 centimetersfrom the bottom of the lamp and the lamp is 12 centimeters deep. How wide is the lamp?? cm? cma(-a, 12)ca. 12)B12 cm| (0.3)13 cmLightsource(0,0)3 cm12 cmcmCheckSave For LaterSubmit Assignment
Answers
It is important to know that the definition of the parabola is
\(x^2=4py\)Where p represents the focus coordinate, so p = 3. Also, we know that y = 12, let's find x
\(\begin{gathered} x^2=4\cdot3\cdot12 \\ x=\sqrt[]{144} \\ x=12 \end{gathered}\)But, the wide of the lamp is 2x, so
\(2x=2\cdot12=24\)Hence, the wide of the lamp is 24 cm.
(6,3); (1,4)
Slope
Giving brainless answer
Answers
Answer:
-1/5. When you do y^2 - y^1/ x^2 - x^1 = It will lead to -1/5
Step-by-step explanation:
MaTh whoever answers this get BrAINLY
Answers
Answer:
addition
-5x-6= -8
-5×= -2
x=2/5
Answer:
addition
Step-by-step explanation:
-5x-6= -8
-5×= -2
x=2/5
What is the sum of 7 x 10-8 and 2 x 10-8 ?
Answers
Answer: 1. \(70x-8\) <----(answer for 7 x 10 - 8)
2. \(20x-8\) <---- (answer for 2 x 10-8)
Step-by-step explanation:
\(7x10\)
\(-8\)
\(2x\)
\(10-8\)
Gina Wilson unit 5 : homework 6: Slope-Intercept Form
Answers
The slope-intercept form of the equation of the line is y = x + 6.
What is an equation of the line?A linear equation with a degree of one is referred to as an equation of the line. Two variables, x, and y, are present in the equation of the line. The third parameter is the line's slope, which indicates the line's elevation.
The equation of the line's general form is:
y = mx + c
m = slope
c = y-intercept
Slope = ( y₂ - y₁ ) / ( x₂ - x₁ )
The formula for calculating the line's slope is
Slope = ( 6 - 4) / ( 2 - 0)
Slope = 1
The y-intercept will be calculated as:-
y = mx + c
y = x +c
6 = c
The equation of the line can be written as:-
y = x + 6
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help me please as soon as possible. I will mark you as the brainliest
Answers
1) 36
2) 143
3) 18
4) 12
5) 15
6) 108
7) 58
The key here is to perform direct substitution in each problem.
I'll do a quick working down here
1) 9(8-4)= 9*4 =36
2) 4(5*4)+7*9= 80+63 =143
3) 114-(3*8*4)= 114-96=18
4) 24/(1*2) = 24/2 = 12
5) 19 - (28/7) = 19- 4 = 15
6) 3 (6)^2 = 3*6*6= 108
7)49 - 3(3)^2 + (6)^2 = 49-27+36= 58
2. 4(5)(4)+7(9)= 80+63= 143
3. 114-3(8)(4)= 114-96= 18
4. 24/(2)(1)= 24/2= 12
5. 19-28/7= 19-4= 15
6. 3(6)^2= 18*18= 324
7. 49-9^2+ 6^2= 68
The speed of light is 299,792,458 meters per second. About how far can a light beam travel in 3 seconds? Write your answer as a product of a single digit and a power of 10.
The light beam can travel about :
meters.
Answers
Answer:
9 x 10 to the power of 8 meters in 3 seconds
Step-by-step explanation:
nine more than three times a number is fifteen
Answers
Answer:
3*2 = 6+9 = 15
Step-by-step explanation:
The required number is 2.
What is an expression?Expressions in maths are mathematical statements that have a minimum of two terms containing numbers or variables, or both, connected by an operator in between.
Given that, nine more than three times a number is fifteen
Let the number be a
Converting the statement into mathematical expression,
9+3a = 15
3a = 6
a = 2
Hence, The required number is 2.
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(If tan A=)
PLZ ANSWER IT FAST...
Answers
Answer:
See explanation
Step-by-step explanation:
\(tan \: A = \frac{2 \sqrt{a} }{a - 1} ...(given) \\ \because \: {sec}^{2} \: A = 1 + {tan}^{2}\: A \\ \therefore\: {sec}^{2} \: A = 1 + \bigg(\frac{2 \sqrt{a} }{a - 1} \bigg)^{2} \\\\ \hspace{38 pt} = 1 + \frac{4a}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{(a - 1)^{2} + 4a}{ {(a - 1)}^{2} } \\ \\\hspace{38 pt} = \frac{a^{2} - 2a + 1+ 4a}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{a^{2} + 2a+ 1}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{(a + 1)^{2}}{ {(a - 1)}^{2} } \\ \\ \therefore {sec}^{2} \: A = \bigg(\frac{a + 1}{ {a - 1}} \bigg) ^{2} \\ \\ \therefore \: {sec} \: A = \pm \bigg(\frac{a + 1}{ {a - 1}} \bigg)\\ \\ \)
In the question It is not mentioned that in which quadrant does angle A lie, so we will assume it to be in first quadrant.
\( \therefore \: {sec} \: A = \bigg(\frac{a + 1}{ {a - 1}} \bigg)\\ \\
\red{ \boxed{ \bold{\therefore \: {cos} \: A = \bigg(\frac{a - 1}{ {a + 1}} \bigg)}}} \\ \\ {sin} \: A ={cos} \: A \times {tan} \: A \\ \\ \hspace{25 pt}=\bigg(\frac{a - 1}{ {a + 1}} \bigg) \times \frac{2 \sqrt{a} }{a - 1} \\ \\ \purple {\boxed { \bold{{sin} \: A = \frac{2 \sqrt{a} }{a + 1}}}}\)