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sammy [17]
3 years ago
12

Walker reads 72 pages of his favorite books in 3 hours. If he reads at a contestant rate, how many pages can he read in an hour?

Mathematics
2 answers:
lara31 [8.8K]3 years ago
8 0

Answer:

24

Step-by-step explanation:

72/3=24 pages per hour

damaskus [11]3 years ago
4 0

Answer:

24 pages

Step-by-step explanation:

It is given that;

72 pages : 3 hours

One must solve for the unit rate of, how many pages per hour. This can be done by dividing both sides of the equation by 3, that way the unit for the parameter (hours) will be reduced to one. Thus, one gets the answer to the question.

72 pages : 3 hours\\/3\\\\24pages : 1 hour

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Simplify 2x(5x + 3) + 7(5x + 3)
zubka84 [21]
Use distributive property to solve.
2x(5x+3) + 7(5x+3)
1. distribute

2x•5x = 10x^2
2x•3= 6x

10x^2+6x

7•5x = 35x
7•3= 21

35x+21

2. combine like terms

10x^2+6x+35x+21

10x^+ 41x +21 [final answer]

6 0
3 years ago
How do you graph quadratic functions
NARA [144]

Answer:

you can plug it into a graphing calcuator right? or no

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
Calculate the average rate of change over the interval 2
Morgarella [4.7K]

Answer:

Average rate of change over the interval 2<= x <= 5:

y = 3x + 5:        3

y = 3x^2 + 1:     21

y = 3^x:             78

<u />

Step-by-step explanation:

2<= x <= 5

Average rate of change over the interval 2<= x <= 5:

<u>y = 3x + 5</u>

y(5) = 3(5) + 5 = 20

y(2) = 3(2) + 5 = 11

Average rate of change = (20 - 11)/(5-2) = 9/3 = <u>3</u>

<u />

<u>y = 3x^2 + 1</u>

y(5) = 3(5^2) + 1 = 75 + 1 = 76

y(2) = 3(2^2) + 1= 13

Average rate of change = (76 - 13)/(5-2) = 63/3 = <u>21</u>

<u />

<u>y = 3^x</u>

y(5) = 3^5 = 243

y(2) = 3^2 =9

Average rate of change = (243-9)/(5-2) = 234/3 =<u> 78</u>

8 0
3 years ago
A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet
Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

L_{1} = 2W_{1}

Also, the area of the walkway of 2 feet wide is 448:

W_{2} = 2 ft

A_{T} = W_{2}*L_{2} = 448 ft^{2}

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

A_{T} = A_{1} + A_{w}    (1)          

The area of the pool is given by:

A_{1} = L_{1}*W_{1}        

A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}  (2)          

And the area of the walkway is:

A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}    (3)          

Where the length of the bigger rectangle is related to the lower rectangle as follows:                  

L_{2} = 4 + L_{1} = 4 + 2W_{1}   (4)        

By entering equations (4), (3), and (2) into equation (1) we have:

A_{T} = A_{1} + A_{w}

A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}                

448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}            

224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}

224 = W_{1}^{2} + 8 + 6W_{1}

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

L_{1} = 2W_{1} = 2*12 ft = 24 ft

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!                                                                                          

8 0
3 years ago
Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] &lt; y0 &lt; [infinity](a) Sketch the graph
kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of  f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0  and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane  is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of  f(y) versus y

Looking at the given differential equation

       \frac{dy}{dt} = e^{y} - 1 for -∞ < y_{o} < ∞

 We can say let \frac{dy}{dt} = f(y) =e^{y} - 1

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

   To fin the critical point we have to set   \frac{dy}{dt} = 0

       This means e^{y} - 1 = 0

                          For this to be possible e^{y} = 1

                          which means that  e^{y} = e^{0}

                          which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0  hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where  f(y) = 0.

In each of the intervals bounded  by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

      This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For  below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper  

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium  that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

5 0
3 years ago
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