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3 years ago

Figure ABCD is a trapezoid. Find

Mathematics

1 answer:

Rudiy273 years ago

3 0

Answer:

x = 3

Step-by-step explanation:

The midsegment is half the sum of the parallel bases, that is

$\frac{1}{2}$ (3x + 3 + 4x + 2) = 13

$\frac{1}{2}$ (7x + 5) = 13 ( multiply both sides by 2 )

7x + 5 = 26 ( subtract 5 from both sides )

7x = 21 ( divide both sides by 7 )

x = 3

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Complete this statement:<br> 15a2x2 + 12a2x + 9a2 = 3a?(

g100num [7]

Answer:

a=0

Step-by-step explanation:

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4 years ago

HELP ME PLEASE ITS THE LAST QUESTION ON THIS TEST

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Step-by-step explanation:

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3 years ago

Which pairs of polygons are congruent?

VashaNatasha [74]

Answer:

Congruent figures are the same size and the same shape.

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4 0

3 years ago

Read 2 more answers

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

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

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

2y+18=42 what is y, and how would you find it?

Nesterboy [21]

Answer:

y=12

Step-by-step explanation:

Use order of operations. Subtract 18 from both sides and the divide by 2.

2y+18 -18=42-18

2y=24

y=12

3 0

3 years ago