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ElenaW [278]
3 years ago
14

The shape of the space is a rectangle. What is the actual area of the space they are filling with cement?

Mathematics
1 answer:
Leni [432]3 years ago
4 0

Answer:

The area would be the length of the rectangle times the with of the rectangle.

Step-by-step explanation:

Imagine the rectangle is cut up into tiny squares. For example, if the length of the rectangle is 5, and the with is 6, than each square would represent 1 unit that is in the area of the rectangle. So if you put 6 squares on the length of that rectangle, that side would be filled up, because the with is 6 and the length would be 4. Add 6 more squares on the with and you have 3 more square colluoms. Add 3 more times 6 than the rectangle would be filled with squares. What we just did is finding the area, which can be simplified to making 5 rows of 6 unit square colloums. This can be even more simplified by the equation length times with

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Answer:

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SOLVE THE QUESTION BELOW ASAP
qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have    

h(t)=-16t^{2}+98

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts,  and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

so

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

h(t)=-16(0)^{2}+98

h(0)=98

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

-16t^{2}+98=0

t^{2}=\frac{98}{16}

square root both sides

t=\pm\frac{\sqrt{98}}{4}

t=\pm7\frac{\sqrt{2}}{4}

therefore

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(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)

(-2.475,0), (2.475,0)

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The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

so

x=0 ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point (0,98)

The y-intercept is the point (0,98)

The t-intercepts are (-2.475,0), (2.475,0)

The axis of symmetry is the y-axis

The graph in the attached figure

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we know that

For t=0

h(t)=-16(0)^{2}+98

h(0)=98\ ft

For t=1

h(t)=-16(1)^{2}+98

h(1)=82\ ft

Find the difference

98\ ft-82\ ft=16\ ft

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

h(t)=-16(1)^{2}+98

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For t=2

h(t)=-16(2)^{2}+98

h(2)=34\ ft

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so

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from  t=0 to t=1

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Answer:

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Answer:

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