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3 years ago

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A scientist builds a Mars rover that can travel 4.9 centimeters per second on flat ground. If the rover continues at the same ra

te, how many centimeters will it travel in 15 1/2 seconds?

2 answers:

larisa [96]3 years ago

8 0

Answer:

The rover would travel 75.95 centimeters in 15 1/2 seconds.

Step-by-step explanation:

Simply multiply 4.9 by 15 1/2.

Hope it helps!

gavmur [86]3 years ago

7 0

Answer:

75.95 cm

Step-by-step explanation:

We know that distance = rate times time

d =rt

We know the rate = 4.9 cm/s

and time = 15 1/2 s = 15.5 s

Substituting these into d=rt

d = (4.9 cm/s) * 15.5 s = 75.95 cm

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Sati [7]

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3 years ago

Read 2 more answers

Suppose you need to know an equation of the tangent plane to a surface S at the point P(4, 1, 4). You don't have an equation for

Alex777 [14]

Answer:

$15x-8y+9z=88$

Step-by-step explanation:

We are given that

Equation of curves

$r_1(t)=$

$r_2(u)=$

Both curves lie on S.

We have to find the equation of tangent plane at P(4,1,4).

$r'_1(t)=,r'_2(u)=$

$r_1(0)=, r_2(1)=$

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Substitute the values in the derivatives

$r'_1(0)=,r'_2(1)=$

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The equation of tangent at point P(4,1,4) is given by

$30(x-4)-16(y-1)+18(z-4)=0$

$30x-120-16y+16+18z-72=0$

$30x-16y+18z=176$

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5 0

3 years ago

BRAINLEST to whoever answer correctly FIRST

lions [1.4K]

It’s B correct answer

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

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2 years ago

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mihalych1998 [28]

This is a good question... but is there multiple choice, i want to know if my answer matches

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2 years ago