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ch4aika [34]
3 years ago
15

Can someone help me with these two problems please and thank you I need them asap

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
8 0
Can please send a bigger pic if you really want someone to solve that problem
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I guess since you have the correct answer there that you need to know how to get it. Start by rewriting that exponent as a radical. It would look like this: \sqrt{(8x-8)^3}=64. The first step is to undo that square root by squaring both sides. That will give you this: (8x-8)^3=4096. Next we have to undo that cubing by taking the cubed root of both sides, leaving us with this: 8x-8=16. We add 8 to both sides to get 8x = 24 and x = 3.

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MA_775_DIABLO [31]

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Read 2 more answers
A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time
user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

a(t) = 24t + 2

Velocity:

v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

v(t) = 12t^{2} + 2t + 13

Position:

s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K

Since s(0) = 3

s(t) = 4t^{3} + t^{2} + 13t + 3

What is its position at time t=5?

This is s(5).

s(t) = 4t^{3} + t^{2} + 13t + 3

s(5) = 4*5^{3} + 5^{2} + 13*5 + 3

s(5) = 593

It's position at time t = 5 is 593.

3 0
3 years ago
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