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3 years ago

Find the additive inverse of -5.3

Mathematics

1 answer:

Fynjy0 [20]3 years ago

3 0

Answer: The additive inverse would be 5.3

Step-by-step explanation: All you do is do the opposite of the number. For example, if the number was -2, the additive inverse would be 2.

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The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

so that

$v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du$

$\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}$

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is $s(0)=0\,\mathrm m$ . Then its displacement over the time interval [0, 3] is

$s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}$

c) The total distance traveled is the integral of the absolute value of the velocity function:

$s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt$

$v(t)$ for $0\le t$ and $v(t)\ge0$ for $\cos^{-1}\left(-\frac29\right)\le t\le3$ , so we split the integral into two as

$\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt$

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

$\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}$

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Step-by-step explanation:

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Write (2p)^3 without exponents

Artemon [7]

For this case we have the following expression:

$(2p) ^ 3$

By properties of exponents we have:

$2 ^ 3p ^ 3$

Then, rewriting the expression we have:

$8p ^ 3$

Since we must write the expression without exponents, then we must remember that:

"In the multiplication, if we have the same base, the exponents are added"

Therefore, by rewriting the expression we have:

$8ppp$

Answer:

The expression without exponents can be written as:

$8ppp$

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