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3 years ago

Will mark Brainlest help plsssss

Mathematics

1 answer:

kherson [118]3 years ago

6 0

Answer:

Step-by-step explanation:

f(0) = $0^{3} +1=1$

f(-2)= $(-2)^{3} +1=-7$

f(x+1) = $(x+1)^{3} +1=x^{3} +3x^{2} +3x+2$

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Step-by-step explanation:

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3 years ago

Calculate the force created from an object with a mass of 1250 grams and an acceleration of 3 m/s^2

abruzzese [7]

The force created by the object is 3.75 Newton.

Given

Mass of the object (m) = 1250 grams

Convert 1250 grams to kilograms by dividing it by 1000.

Mass (m) = 1.25 kg

Acceleration of the body (a) = 3 $m/s^{2}$

From Newton's second law we know $F= m *a$

By Substituting in the above equation we get

Force = 1.25 x 3

Force = 3.75 $kg/m /s^{2}$

It can also be written as force = 3.75 N

Hence Force created by the body is 3.75 Newton.

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1 year ago

Jacy has a collection of 140 coins worth $19.00. She has only nickels, dimes, and quarters. The difference in the

Colt1911 [192]

Answer:50(b)

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$