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2 years ago

15

Your grade at the start of quarter 3 was a 68, at the end of the quarter it was a 91. What is the percent change in grade?

1 answer:

Georgia [21]2 years ago

7 0

91 - 68 = 23
23/91 x 100= 25.3%

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3 years ago

(-4,-6) and (3,-7) find the distance between each pair of points

Vlad [161]

Answer:

$5\sqrt{2}$

Step-by-step explanation:

distance between two points:

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}$

we have:

(-4, -6), (3, -7)

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so we have:

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

7 0

3 years ago