pls help asap!!☹️

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The angle measures of the triangles before and after dilation are the same

Given that, we have a triangle that is dilated to form another triangle by a scale factor of n

The dilation transformation is a rigid transformation

This means that it changes the size of a shape after it is applied

However, the shape and the image would be similar shapes and as such would have their angles unchanged

This means that irrespective of the value of the scale factor n, the angle measures would remain the same

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Sin (30) = 1/2

Option B is the correct answer.

Step-by-step explanation:

\( \sin(30) = \frac{1}{2} \)

Answer:

From what I found, the answer is parallel

Step-by-step explanation:

not much of an explanation haha

The graph of the system of equations, 2x + y = 6  and 6x + 3y = 12 is:  C. Intersecting lines

Note:

Given that 2x + y = 6  and 6x + 3y = 12 is a system of equations, when both equations are graphed, it will give us two intersecting lines.

The coordinate of the point of intersection is the solution to the system of equations.

Therefore, the graph of the system of equations, 2x + y = 6  and 6x + 3y = 12 is:  C. Intersecting lines

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Answer:

The formula for the volume of a cone is

V = (1/3)πr²h.

Answer:

(-10,3),(9,0)

Step-by-step explanation:

Answer: y= x + 4

Step-by-step explanation:

Answer:

answer is 15 and second one i dont know yet sorry am doing scoo

Step-by-step explanation:

To stop a 1200 kg car moving at 99 km/h in a straight path,

the work that must be done in Joules is 453750.

Step-by-step Given,

Mass of car, m = 1200 kg

Initial velocity, u = 99 km/h = 27.5 m/s

Final velocity, v = 0 m/s (as the car is brought to rest)

Initial kinetic energy, K.E1 = 1/2 m u²

Final kinetic energy, K.E2 = 1/2 m v²

Work done to bring the car to rest,

W = K.E1 - K.E2W

= 1/2 m u² - 1/2 m v²W

= 1/2 × 1200 × (27.5)² - 1/2 × 1200 × (0)²W

= 1/2 × 1200 × (27.5)²W

= 453750 J

Therefore, the work that must be done in Joules to stop the car is 453750.

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Where is the table?

The Maclaurin series for f(x) converges absolutely for x within the interval (-2/3, 2/3).

To find the Maclaurin series for the function f(x) = (3cos(x))ln(1+x), we can use the standard formulas for the Maclaurin series expansion of elementary functions.

First, let's find the derivatives of f(x) up to the third order:

f(x) = (3cos(x))ln(1+x)

f'(x) = -3sin(x)ln(1+x) + (3cos(x))/(1+x)

f''(x) = -3cos(x)ln(1+x) - (6sin(x))/(1+x) + (3sin(x))/(1+x)² - (3cos(x))/(1+x)²

f'''(x) = 3sin(x)ln(1+x) - (9cos(x))/(1+x) + (18sin(x))/(1+x)² - (12sin(x))/(1+x)³ + (12cos(x))/(1+x)² - (3cos(x))/(1+x)³

Next, we evaluate these derivatives at x = 0 to find the coefficients of the Maclaurin series:

f(0) = (3cos(0))ln(1+0) = 0

f'(0) = -3sin(0)ln(1+0) + (3cos(0))/(1+0) = 3

f''(0) = -3cos(0)ln(1+0) - (6sin(0))/(1+0) + (3sin(0))/(1+0)² - (3cos(0))/(1+0)² = -3

f'''(0) = 3sin(0)ln(1+0) - (9cos(0))/(1+0) + (18sin(0))/(1+0)² - (12sin(0))/(1+0)³ + (12cos(0))/(1+0)² - (3cos(0))/(1+0)³ = -9

Now we can write the first three nonzero terms of the Maclaurin series:

f(x) = f(0) + f'(0)x + (f''(0)/2!)x² + (f'''(0)/3!)x³ + ...

f(x) = 0 + 3x - (3/2)x² - (9/6)x³ + ...

Simplifying, we have:

f(x) = 3x - (3/2)x² - (3/2)x³ + ...

To determine the values of x for which the series converges absolutely, we need to find the interval of convergence. In this case, we can use the ratio test:

Let aₙ be the nth term of the series.

|r| = lim(n->infinity) |a_(n+1)/aₙ|

= lim(n->infinity) |(3/2)(xⁿ+1)/(xⁿ)|

= lim(n->infinity) |(3/2)x|

For the series to converge absolutely, we need |r| < 1:

|(3/2)x| < 1

|x| < 2/3

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The triple integral of x²y²z² over the given limits is evaluated by integrating with respect to x, y, and z is -243.

Firstly, we integrate x² from π to 0, which gives -8/3. Then, we integrate y² from 0 to 3, which gives 27/3. Finally, we integrate z² from 0 to 3, which gives 27/3. Multiplying all the values together, we get the final answer of -243. Therefore, the value of the given triple integral is -243.

In order to evaluate a triple integral, we must integrate the given function over three variables with respect to their limits of integration. In this case, we are integrating x²y²z² over the limits 0≤x≤2, 0≤y≤3, and 0≤z≤3. We start by integrating with respect to x, then y, and finally z, using the limits given.

Each integral gives us a value which is then multiplied together to get the final answer. It is important to follow the order of integration correctly, and to keep track of the limits for each variable. In this way, we can evaluate the triple integral and find the value of the given function over the given region of integration.

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Answer:

30

Step-by-step explanation:

1. x + 40 = 3x - 20

2. 40 = 2x - 20

3. 60 = 2x

4. 30 = x

Answer:

No Solution

Step-by-step explanation:

\(\tt -2\left(4x-1\right)=-8x-1\)

Expand:-

\(\tt -8x+2=-8x-1\)

Add 8x to both sides:-

\(\tt -8x+2+8x=-8x-1+8x\)

Simplify:-

\(\tt 2=-1\)

The sides are not equal, therefore there is No Solution.

_____________________

Hope this helps!
Have a nice day!

Answer:

He would pay $11.50 for two pounds of walnuts with the coupon.

7x - 2.5

x is the pounds of walnuts purchased

Answer:

1) C

Step-by-step explanation:

Step 1)

First we have to plot ↓

2x ≥ y + 2

And that looks like Image 1

Step 2)

Second we have to plot ↓

2x - 5y ≤ 10

and that looks like Image 2

To determine the level of service of the four-lane freeway, we need to consider the geometric design, traffic volume, vehicle mix, and the presence of ramps and obstructions. The HCM methodology can be used to estimate the level of service based on various parameters.

To determine the level of service of the four-lane freeway, we need to consider several factors, including the geometric design, traffic volume, vehicle mix, and the presence of ramps and obstructions.

First, the four-lane freeway has two lanes in each direction with 12-ft lanes, and obstructions 3 ft from the right edge of the traveled pavement. This geometric design may affect the traffic flow, especially when heavy vehicles and RVs are present. These types of vehicles require more space to maneuver, and the presence of obstructions may limit their ability to do so.

Second, the directional weekday peak-hour volume of 2100 vehicles is observed. This traffic volume is relatively high and may lead to congestion and delay, especially during peak hours. To estimate the level of service, we need to use the Highway Capacity Manual (HCM) methodology, which considers various parameters such as speed, travel time, and density.

Third, we need to account for the vehicle mix, which includes 5% heavy vehicles and 10% RVs. These types of vehicles have different characteristics than passenger cars and may affect the traffic flow differently. For example, heavy vehicles may cause more congestion due to their slower speed and larger size, while RVs may require more space to maneuver and may have limited visibility.

Finally, we need to consider the presence of ramps within 2 miles upstream and 2 miles downstream of the midpoint of the analysis segment. Ramps can affect the traffic flow by adding or removing vehicles from the mainline, and their location and design can affect the merging and weaving maneuvers.

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Answer:

10 1/2 pounds are needed.

Step-by-step explanation:

7 x 1.5 = 10.5

The correct inequality is:  C) 2x + 5 > -1.

Given that, the solution set of an inequality is graphed on the number line below.  { -4, -3, -2, -1, 0, 1}.

Looking at the solution set, observe that all the values are less than or equal to 1.

The solution sets for each inequality:

A) 2x + 5 < -1:

Subtracting 5 from both sides:

2x < -6

Dividing both sides by 2:

x < -3

The solution set is (-∞, -3).

B) 2x + 5 > -1:

Subtracting 5 from both sides:

2x > -6

Dividing both sides by 2:

x > -3

The solution set is (-3, +∞).

C) 2x + 5 > -1:

Subtracting 5 from both sides:

2x > -6

Dividing both sides by 2: x > -3

The solution set is (-3, +∞).

D) 2x + 5 > -1 + 2:

Simplifying the right side:

2x + 5 > 1

Subtracting 5 from both sides:

2x > -4

Dividing both sides by 2: x > -2

The solution set is (-2, +∞).

Therefore, the solution sets are:

A) Solution set: (-∞, -3),

B) Solution set: (-3, +∞)

C) Solution set: (-3, +∞)

D) Solution set: (-2, +∞).

Hence, the correct inequality is:  C) 2x + 5 > -1.

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The relation 2 represents a function.

In order to determine which relation in the below table represents a function, we need to first understand what a function is.A function is a relationship in which each input value corresponds to exactly one output value.

To put it another way, each x-value has one and only one y-value. The most typical method to determine whether a relation is a function is to use the vertical line test.

The vertical line test is a way to determine if a relation is a function graphically. To test if a graph is a function, we draw a vertical line through each x-value on the graph. If a vertical line crosses the graph more than once, it is not a function.

If, on the other hand, the graph passes the vertical line test and no vertical line crosses the graph more than once, it is a function.Now let's look at the table below to determine which relation is a function.

We will first plot the x and y values of each relation on a coordinate system and then apply the vertical line test to each relation.

Relation 1: x | y0 | 10 | 11 | 22 | 23 | 34 | 35 | 4Relation 1 does not represent a function since we can draw a vertical line through x = 3 and the line will cross the graph more than once.

Relation 2: x | y2 | 33 | 34 | 45 | 46 | 57 | 5Relation 2 represents a function since we can draw a vertical line through each x-value on the graph and it will only cross the graph once.

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Answer:

3

Step-by-step explanation:

15g

g=1/5

15×1/5

15/1×1/5=3

Answer:

3

Step-by-step explanation:

This is the answer because:

1) We have to substitute the value into the equation:

15(1/5)

2) Next, we have to mutiply 15 and 1/5 together since the g is just a variable that is supposed to be mutiplied with 15:

15(1/5) = 15/5 = 15 ÷ 5 = 3

Therefore, the answer is 3.

Hope this helps! :D

Answer:

Friday & Sunday

Step-by-step explanation:

Thursday: 287

Friday: 618

Saturday: 973

Sunday: 532

973 < 287 + 618 = 973 < 905

We now know this isn't true.

973 < 287 + 532 = 973 < 819

This isn't true either.

973 < 532 + 618 = 973 < 1150

This is true.  The two days with the time are Friday and Sunday.

I really hope this helps you!  Tell me if it's right or not! :D

The sample data range is = 2.68

From the question, we have the  grade-point averages apply to a random sample of graduating seniors are:

3.88 , 2.73, 2.71, 3.09, 3.28, 3.51, 2.86, 1.20, 3.13, 3.24

As we know that:

The range of the data is calculated by using the following formula;

Range = Highest value in the data - Lowest value in the data

And in the given data,

Highest value in the data is = 3.88

Lowest value in the data is = 1.20

Now, Put all the values in above formula:

Range = 3.88 - 1.20

Range = 2.68

So, the sample data range is = 2.68

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Answer:

relax jamal dont pull up the 9

Step-by-step explanation:

Based on the data, it can be inferred that Emily wire will cost 13.8 and Andy wire will cost 25.2

To calculate the value of the wire for Andy and Emily we must take into account the conversion factor of inches, feet, and yards.

1 = 12 ℎ

1 = 3 = 36 ℎ

According to the above, we must find the number of inches they need and identify the cost that it will have according to the information in the table.

Andy needs 23 feet

Enily needs 14 yards

According to the above, we must calculate how much the wire will cost each one as shown below:

20 feet of wire are worth $1. So, we must make a rule of three to identify the price of each wire:

Note: This question is incomplete. Here is the complete information

Image attached

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Answer:

The answer is 2022537/290000

problem:

The three-point formula of order O(h²) to approximate f'(zo) using f(xo - h), f(xo), and f(xo + 2h) is derived as follows:

To approximate the derivative f'(zo) using a three-point formula, we can use Taylor series expansions. Let's consider the Taylor expansions of f(xo - h), f(xo), and f(xo + 2h) around xo:

f(xo - h) = f(xo) - hf'(xo) + (h²/2)f''(xo) - (h³/6)f'''(xo) + O(h⁴)

f(xo) = f(xo)

f(xo + 2h) = f(xo) + 2hf'(xo) + (4h²/2)f''(xo) + (8h³/6)f'''(xo) + O(h⁴)

Now, we can combine these expansions to eliminate the second derivative term and obtain an approximation for f'(zo). Rearranging the equations and solving for f'(xo), we get:

f'(xo) ≈ (1/2h) [4f(xo) - 3f(xo - h) - f(xo + 2h)]

This three-point formula has an error term of O(h²), indicating that the approximation improves quadratically as the step size h decreases. By using this formula, we can estimate the derivative f'(zo) based on function evaluations at three points: f(xo - h), f(xo), and f(xo + 2h).

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