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4 years ago

12

Goes along with page 1 another page coming

1 answer:

Salsk061 [2.6K]4 years ago

6 0

There you go ^-^ Your intersection point is at (-3, 4). Good luck :3

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Jake tosses a coin up in the air and lets it fall to the ground. The equation that models the height (in feet) and time (in seco

adoni [48]

Answer:

The maximum height of coin Jake tosses is 15 feet.

Step-by-step explanation:

Given Jake tosses a coin up in the air

and the height of coin is model by h(t)= $(-16)t^{2} +24t+6$

To find height of the coin when Jake tosses it:

When a coin is in the hand of Jake, time t=0

Height of coin is h(t)= $(-16)t^{2} +24t+6$

h(0)= $(-16)(0)^{2} +24(0)+6$

h(0)=6 feet.

Therefore, Height of coin at time t=0 is 6.

For maximum height of the coin,

h(t)= $(-16)t^{2} +24t+6$

Differentiating both side,

$\frac{d}{dt}h(t)=\frac{d}{dt}[(-16)t^{2} +24t+6]$

$\frac{d}{dt}h(t)=\frac{d}{dt}[(-32)t+24]$

$\frac{d}{dt}h(t)=0$

$\frac{d}{dt}[(-32)t+24]=0$

t= $\frac{24}{32}$

t= $\frac{3}{4}$

t=0.75

Now,

h(t)= $(-16)t^{2} +24t+6$

h(0.75)= $(-16)(0.75)^{2} +24(0.75)+6$

h(0.75)=15 feet.

The maximum height of coin jake tosses is 15 feet.

7 0

4 years ago

Combine like terms. 8x-x

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[tex]8x - x \\ 8 (0)

hope this help
brainliest plz

4 0

3 years ago

Read 2 more answers

At a local hardware store, Ms. Jones learned that 1 would cost her $.50, 12 would cost $1.00, and the price of 144 would be $1.5

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House/garage numbers? Like what you stick on your house so the postman knows which one is which.

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3 years ago

An object dropped from rest from the top of a tall building on Planet X falls a distance feet in the first t seconds. Find the a

ch4aika [34]

Complete Question:

An object dropped from rest from the top of a tall building on Planet X falls a distance d(t)=13t^2 feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from t1=2 to t2=10. This rate is known as the average velocity, or speed.

The average velocity as t changes from 2 to 10 seconds is _______ feet/sec?

Answer:

$v = 156$

Step-by-step explanation:

Given

$d(t)=13t^2$

Time Interval: (a,b)

$a = 2$

$b = 10$

Required

Determine the average rate of change (v)

This is calculated as thus:

$v = \frac{d(b) - d(a)}{b - a}$

Substitute values for b and a

$v = \frac{d(10) - d(2)}{10 - 2}$

$v = \frac{d(10) - d(2)}{8}$

Calculate d(10): Substitute 10 for t

$d(t)=13t^2$

$d(10) = 13 * 10^2$

$d(10) = 13 * 100$

$d(10) = 1300$

Calculate d(2): Substitute 2 for t

$d(t)=13t^2$

$d(2) = 13 * 2^2$

$d(2) = 13 * 4$

$d(2) = 52$

The equation $v = \frac{d(10) - d(2)}{8}$ becomes

$v = \frac{1300 - 52}{8}$

$v = \frac{1248}{8}$

$v = 156$

Hence, the average rate of change is 156ft/s

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3 years ago

solve this system of linear equations separate the x and y values with a comma -14x=-7-7y -10x=-15-7y

gizmo_the_mogwai [7]

X=7/4+7/4y ,y∈ (One symbol I can't find but that's what I got)

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3 years ago

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