The volume of a cube that is 2.9 x 5 x 4

Question 1:

(a) The equation representing Elaine's total parking cost is:

C = x * t

(b) So the cost of parking for a full 24 hours would be 24 times the cost per hour.

Question 2:

The given system of equations is inconsistent and has no solution.

(a) To represent Elaine's total parking cost, C, in dollars for t hours, we need to know the cost per hour. Let's assume the cost per hour is $x.

(b) If Elaine wants to park her car for a full 24 hours, we can substitute t = 24 into the equation from part (a):

C = x * 24

Question 2:

To solve the linear system:

-x - 6y = 5

x + y = 10

We can use the elimination method.

Multiply the second equation by -1 to create opposites of the x terms:

-x - 6y = 5

-x - y = -10

Add the two equations together to eliminate the x term:

(-x - 6y) + (-x - y) = 5 + (-10)

-2x - 7y = -5

Now we have a new equation:

-2x - 7y = -5

To check the answer, we can substitute the values of x and y back into the original equations:

From the second equation:

x + y = 10

Substituting y = 3 into the equation:

x + 3 = 10

x = 10 - 3

x = 7

Checking the first equation:

-x - 6y = 5

Substituting x = 7 and y = 3:

-(7) - 6(3) = 5

-7 - 18 = 5

-25 = 5

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To prove that there are only these five regular polyhedra, we can consider Euler's polyhedron formula, which states that for any convex polyhedron, the number of vertices (V), edges (E), and faces (F) satisfy the equation V - E + F = 2.

The five regular polyhedra, also known as the Platonic solids, are the only convex polyhedra where all faces are congruent regular polygons, and the same number of polygons meet at each vertex.

The five regular polyhedra are:

1. Tetrahedron: It has four triangular faces, and three triangles meet at each vertex.

2. Cube: It has six square faces, and three squares meet at each vertex.

3. Octahedron: It has eight triangular faces, and four triangles meet at each vertex.

4. Dodecahedron: It has twelve pentagonal faces, and three pentagons meet at each vertex.

5. Icosahedron: It has twenty triangular faces, and five triangles meet at each vertex.

To prove that there are only these five regular polyhedra, we can consider Euler's polyhedron formula, which states that:

"for any convex polyhedron, the number of vertices (V), edges (E), and faces (F) satisfy the equation V - E + F = 2".

For regular polyhedra, each face has the same number of sides (n) and each vertex is the meeting point of the same number of edges (k). Therefore, we can rewrite Euler's formula for regular polyhedra as:

V - E + F = 2  

=>  kV/2 - kE/2 + F = 2  

=>  k(V/2 - E/2) + F = 2

Since each face has n sides, the total number of edges can be calculated as E = (nF)/2, as each edge is shared by two faces. Substituting this into the equation:

k(V/2 - (nF)/2) + F = 2  

=>  (kV - knF + 2F)/2 = 2  

=>  kV - knF + 2F = 4

Now, we need to consider the conditions for a valid polyhedron:

1. The number of faces (F), edges (E), and vertices (V) must be positive integers.

2. The number of sides on each face (n) and the number of edges meeting at each vertex (k) must be positive integers.

Given these conditions, we can analyze the possibilities for different values of n and k. By exploring various combinations, it can be proven that the only valid solutions satisfying the conditions are:

(n, k) pairs:

(3, 3) - Tetrahedron

(4, 3) - Cube

(3, 4) - Octahedron

(5, 3) - Dodecahedron

(3, 5) - Icosahedron

Therefore, there exist only five regular polyhedra.

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The total number of animals are 112 if the ratio of kangaroos to emus is 9:7 and the number of emus are 49.

A ratio in mathematics displays the multiplicative relationship between two numbers. In a bowl of fruit, for instance, the proportion of oranges to lemons is eight to six if there are eight oranges and six lemons (that is, 8:6, which is equivalent to the ratio 4:3).

A ratio can have any number of numbers as its components, including counts of people or things, weights, lengths, and times, among other values.

Given,

The ratio of kangaroos to emus is 9:7

And also given that the number of emus=49

Since, the ratio is 9:7

The number of kangaroos=63

63:49

9:7

Total number of animals=63+49

=112

Therefore, the total number of animals are 112 if the ratio of kangaroos to emus is 9:7 and the number of emus are 49.

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3/5 or 60% of the prism's volume will be filled with sand.

The volume of each cube is e³ = 5³ = 125 cubic centimeters.

Since the rectangular prism is made up of two cubes, its volume is 2e³ = 2(125) = 250 cubic centimeters.

If Arusha fills the prism with 150 cubic centimeters of sand, the fraction of the prism's volume that will be filled with sand is:

150/250 = 3/5

Therefore, 3/5 or 60% of the prism's volume will be filled with sand.

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The probability that Aakesh will select a blue marble from each bag is 4/45.

To find the probability that Aakesh will select a blue marble from each bag, we need to calculate the probability of selecting a blue marble from each bag and then multiply those probabilities together.

Let's start with the first bag, which contains 5 marbles: 2 red marbles, 2 blue marbles, and 1 green marble. The probability of selecting a blue marble from the first bag is:

P(Blue from first bag) = Number of blue marbles / Total number of marbles in the first bag

P(Blue from first bag) = 2 / 5

Now, let's move on to the second bag, which contains 9 marbles: 3 red marbles, 2 blue marbles, and 4 green marbles. The probability of selecting a blue marble from the second bag is:

P(Blue from second bag) = Number of blue marbles / Total number of marbles in the second bag

P(Blue from second bag) = 2 / 9

To find the probability of both events happening (selecting a blue marble from each bag), we multiply the individual probabilities together:

P(Blue from both bags) = P(Blue from first bag) * P(Blue from second bag)

P(Blue from both bags) = (2 / 5) * (2 / 9)

P(Blue from both bags) = 4 / 45.

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Answer:

400，000，000，000

Step-by-step explanation:

2*2=4

and 5 zeros plus 6 zeros equals 11 zeros.

Answer: the answer is 400000000000

Step-by-step explanation:

20 times 20 is 40 and we add all zeros

hope this helps!

Answer:

2/10, 0.2, 20%

Step-by-step explanation:

hope this helps

Answer:

Matt needs to read 55 books for the school book fair. He has already read 7 and reads books at a rate of 4 books per day. How many more books will he need to read?

(can I get brainliest pls)

The solution is Option A.

The coordinate of the triangle after the translation is given by A' ( 0 , -3 ) with 4 units right and 4 units down

A translation moves a shape up, down, or from side to side, but it has no effect on its appearance. A transformation is an example of translation. A transformation is a method of changing a shape's size or position. Every point in the shape is translated in the same direction by the same amount.

Given data ,

Let the coordinate of the triangle be represented as A

Now , the coordinate A = A ( -4 , 1 )

Now , the coordinate of the triangle after translation is A' ( 0 , -3 )

when there is a translation of the coordinate A by 4 units to the right , we get

A to 4 units right = A ( -4 + 4 , 1 )

The new coordinate of A = A ( 0 ,  1 )

Now , A to 4 units down , we get

The coordinate of A' = A' ( 0 , 1 - 4 )

The coordinate of A' = A' ( 0 , -3 )

Hence , the translated coordinate is A' ( 0 , -3 )

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Answer:

all numbers are 2X

Step-by-step explanation:

look 5×2=10×2=20×2=40×2=80

Answer:

Step-by-step explanation: The answer for 7: is 7/2

8: 28/5.

9: 9/4

10: 24/5

f cannot have a jump discontinuity at \($x_0 \in(a, b)$\) and  \($$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$\)

f cannot have a removable discontinuity at \($$x_0 \in(a, b) $$\) and \(\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)\)

Let f be a function defined on (a, b) satisfies intermediate value property.

Claim: f ca not have removable on jump discontinuity.

Suppose f has a jump discontinuity at \($x_0 \in(a, b)$\)

We take \($\theta$\) such that

\($$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$\)

Now there exist \($\delta > 0$\) such that \($f(x) < \theta$\) for all \($x \in\left[x_0-\delta, x_0\right)$\) and \($f(x) > \theta$\) for all \($x \in\left(x_0, x_0+\delta\right]$\)

Now \($f\left(x_0-\delta\right)\)\(< \theta < f\left(x_0+\delta\right)$\) for all \($x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$\) and \($f\left(x_0\right) \neq \theta$\).

Therefore the point \($\theta$\) has no preimage under f

that is, there does not exists \($y \in\left[x_0-\delta, x_0+\delta\right]\) for which

\($$f(y)=\theta\) because \(\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$\)

Therefore f does not satisfies intermediate value property on \($\left[x_0-\delta, x_0+\delta\right]$\),

Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),

Therefore f can not have a jump discontinuity.

Suppose f has a removable point of discontinuity at \($x_0 \in(a, b)$\),

Let \($\alpha=\lim _{\alpha \rightarrow x_0} f(x)$\),

Let \(\alpha < f\left(x_0\right)$\) so \($f\left(x_0\right)-\alpha > 0$\).

Now \($\lim _{x \rightarrow x_0} f(x)=\alpha$\) then \(\exists$ \delta > 0$\) such that

\($$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$\)

So \($f(x) < \frac{f\left(x_0\right)+\alpha}{2}$\) for all \($x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$\)

Now \($f\left(x_0\right) > \alpha$\).

And  \($f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$\) for all \($x \in\left[\left(x_0 \delta, x_0\right)\right.$\)

Let \($\mu=\frac{f\left(x_0\right)+\alpha}{2}$\).

Then there does not exist \($e \in\left[x_0-\delta, c\right]$\) such that \($f(c)=\mu$\)

Because for \($e=x_0 \quad f(e) > \mu$\)

                for \($c < x_0 \quad f(c) < \mu$\).

Therefore f does not satisfy IVP on \($\left[x_0-\delta_1 x_0\right]$\) which contradict our hypothesis,

therefore \($\alpha \geqslant f\left(x_0\right)$\)

Let \($\alpha > f\left(x_0\right)$\). so \($\alpha-f\left(x_0\right) > 0$\)

\($\lim _{x \rightarrow x_0} f(x)=\alpha$\)

Then \(\exists $ \varepsilon > 0$\) such that

\($|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$\) for all \($\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$\)

\($\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$\) for all \($x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$\)

\($\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$\) for all \($x \in\left[x_0-\varepsilon_1, x_0\right)$\)

Now \($f\left(x_0\right) < \alpha$\)

The \($f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$\).

So \($f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$\) for all \($x \in\left[x_0 \varepsilon, \varepsilon_0\right)$\)

Let \($\eta=\frac{f\left(x_e\right)+\alpha}{2}$\)

Then there does not exist \($d \in\left[x_0-\varepsilon, x_0\right]$\) such that \($f(d)=\xi$\).

Because if \($d=x_0, f(d)=f\left(x_0\right) < \eta$\) if \($d E\left[x_0-\varepsilon, x_0\right)$\)

Then \($f(d) > \eta$\)

Therefore f does not satisfies IVP on \($\left[x_0-\varepsilon, x_0\right]$\) which contradict olio hypothesis.

Therefore \($\alpha \leq f\left(x_0\right)$\) (b) From (a) and (b) it follows \($\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$\). Therefore f can not have a removable discontinuous

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Answer:(2x+7)+(2x+7)

Step-by-step explanation:

what you do is put 2x on the top and on the side and then seven on the top and on the side and then you got your answer

Start at -10. Note that to get to -16, we need to move 2 spaces. This is a distance of 6 units.

2 spaces = 6 units

1 space = 3 units (divide both sides by 2)

Each tickmark represents 3 units.

Go another 2 spaces over and we go 2*3 = 6 units to the left getting to -16-6 = -22

For the population to double, it takes around 7.07 hours.

The continuous exponential growth model is given by:

\(N(t) = N0 * e^{rt}\)

Where:

N(t) is the population size at time t

N is the initial population size (at t = 0)

r is the growth rate parameter

t is the time elapsed

We want to find the time t it takes for the population size to double, which means N(t) = 2N. Substituting into the growth model, we get:

\(2N = N * e^{rt}\)

Dividing both sides by N, we get:

\(2 = e^{rt}\)

Taking the natural logarithm of both sides, we get:

ln(2) = rt

Solving for t, we get:

t = ln(2)/r

Substituting r = 0.098 (9.8% as a decimal), we get:

t = ln(2)/0.098

t ≈ 7.07 hours

Therefore, it takes approximately 7.07 hours for the population size to double.

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Answer:

there is nothing to look at or answer

Answer:

Step-by-step explanation:

7 + 28 + 7 = 42 units^2     see image below

Answer:

2.3 hours

Step-by-step explanation:

From 3:12 pm to 4:12 pm, one hour passes.

From 4:12 pm to 5:12 pm, another hour passes.

From 5:12 pm to 5:30 pm, there are 18 minutes that pass.

Therefore, the total time elapsed is 2 hours and 18 minutes out of 60 minutes in an hour.

To convert this to a fraction, we can write:

2 hours + 18 minutes = 2 + 18/60 hours = 2.3 hours

So, 2.3/1 hour passes from 3:12 pm to 5:30 pm.

Answer:

2 forms

Step-by-step explanation:

hope this helps <3

\( \sf \longrightarrow \: \frac{8}{5} \div 6 \\ \)

\( \sf \longrightarrow \: \frac{5}{8} \times 6 \\ \)

\( \sf \longrightarrow \: \frac{5 \times 6}{8} \\ \)

\( \sf \longrightarrow \: \frac{30}{8} \\ \)

\( \sf \longrightarrow \: \frac{8}{30} \\ \)

\( \sf \longrightarrow \: 0.26 \\ \)

_____________________________

Answer: C i took the test

" If you buy a new cell phone, your opportunity cost is the time you could spend talking on the phone."

Step-by-step explanation:

I took the test and got an A hope this helps.

The required estimate for the reduced Total Wages for the next two weeks combined would be approximately 1.95 times the amount of money being paid to all employees over the course of a week in the Current Week.

The percentage is the ratio of the composition of matter to the overall composition of matter multiplied by 100.

Here,
To estimate the reduced Total Wages for the next two weeks combined, we can use the following formula:

Reduced Total Wages = Current Total Wages x (1 - Wage Reduction Percentage)

Reduced Total Wages = C × (1 - 0.025) × 2

Simplifying this expression, we get:

Reduced Total Wages = C × 0.975 × 2

Reduced Total Wages = 1.95C

Therefore, the estimate for the reduced Total Wages for the next two weeks combined would be approximately 1.95 times the amount of money being paid to all employees over the course of a week in the Current Week.

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The government agency as a monopolist will produce and sell internet connections up to the point where the marginal cost is 1/2. The price will be set at 1, given the perfectly elastic demand function.

In the scenario where a government agency has a monopoly in the provision of internet connections and the inverse demand function is given by p = 1, we can analyze the market equilibrium and the implications for pricing and quantity.

The inverse demand function, p = 1, implies that the market demand for internet connections is perfectly elastic, meaning consumers are willing to purchase any quantity of internet connections at a price of 1. As a monopolist, the government agency has control over the supply of internet connections and can set the price to maximize its profits.

To determine the optimal pricing and quantity, the monopolist needs to consider the marginal cost of providing internet connections. In this case, the marginal cost is given as 1/2. The monopolist will aim to maximize its profits by equating marginal cost with marginal revenue.

Since the inverse demand function is p = 1, the revenue received by the monopolist for each unit sold is also 1. Therefore, the marginal revenue is also 1. The monopolist will produce up to the point where marginal cost equals marginal revenue, which in this case is 1/2.

As a result, the monopolist will produce and sell internet connections up to the quantity where the marginal cost is 1/2. The monopolist will set the price at 1 since consumers are willing to pay that price.

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Answer:

y = 3x - 7

Step-by-step explanation:

For an equation in the format y = ax + b, a is the slope and b is the y intercept.

Answer:

333

Step-by-step explanation:

9514 1404 393

Answer:

  x = 1/4(y -4)² +7

Step-by-step explanation:

A parabola can have the equation ...

  x = 1/(4p)(y -k)² +h

where the vertex is (h, k), and the vertex-to-focus distance is p.

The vertex is halfway between the focus and directrix, so has coordinates ...

  ((8, 4) +(6, 4))/2 = (7, 4)

The distance from the vertex to the focus is 8-7 = 1, so the equation can be written ...

  x = 1/4(y -4)² +7

By using slope intercept form of equation of line, the graph of

h(x) = \(-\frac{2}{5}x + 4\) has been shown

What is equation of line in slope intercept form?

The most general form of equation of line in slope intercept form is given by y = mx + c

Where m is the slope of the line and c is the y intercept of the line.

Slope of a line is the tangent of the angle that the line makes with the positive direction of x axis.

If  \(\theta\)  is the angle that the line makes with the positive direction of x axis, then slope (m) is given by

m = \(tan\theta\)

The distance from the origin to the point where the line cuts the x axis is the x intercept of the line.

The distance from the origin to the point where the line cuts the y axis is the y intercept of the line.

Let the independent variable be x and dependent variable be h(x)

The dependent variable decreases 2 units for every 5 units the independent variable increase.

Slope = \(-\frac{2}{5}\)

So,

h(x) = \(-\frac{2}{5}x + c\)

Now,

h(0) = 4

So, c = 4

h(x) = \(-\frac{2}{5}x + 4\)

The graph has been attached here-

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The h(x) should be Left 4 units translated to the left or right to represent the same function d(x).

The function is a type of relation, or rule, that maps one input to specific single output.

Calculation of h(x) that need to be translated:

h(x) represents a line with slope 2.  

Let's assume it has y-intercept b so,

h(x) = 2x + b

d(x) represent h(x) shifted up 8 units.  

So,

d(x) = h(x) + 8

d(x) = 2x + b + 8

Now

h(x−a) = d(x)

2(x−a) + b = 2x + b + 8

2x − 2a + b = 2x + b + 8

-2a = 8

a = -4

Since a is negative, so the shift is to the left to represent the same function d(x) .

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