3 years ago

How to solve G(2x-3)= (2x-3+a)-(2x-3+a)

Mathematics

1 answer:

madreJ [45]3 years ago

3 0

Given f (x) = 2x, g(x) = x + 4, and h(x) = 5 – x3, find (f + g)(2), (h – g)(2), (f × h)(2),

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The equation r(t)= (2t)i + (2t-16t^2)j is the position of a particle in space at time t. Find the angle between the velocity and

ankoles [38]

Answer:

The answer is 135 degrees.

Step-by-step explanation:

As we are given the position. If we take the <u>derivative</u>, we get the velocity vector. If we take the <u>derivative</u> again, we find the acceleration vector of the particle.

$r(t)=(2t)i+(2t-16t^{2})j$

$V(t)=2i+(2-32t)j$

$a(t)=-32j$

At time t=0;

$v(t)=2i+2j$

$a(t)=-32j$

As i attach in the picture the angle between the velocity and acceleration vector is $(45+90)=135$ degrees

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3 years ago

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damaskus [11]

Answer:

921

Step-by-step explanation:

If he recives 3x amount than his oppenent

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one

Step-by-step explanation:

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3 years ago

What is the value of the expression below when y=2 6y^2 +8y-9

Annette [7]

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Step-by-step explanation:

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Natalka [10]

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2/3.......... :)

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