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2 years ago

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A group of tourists spends $168 to rent snorkels and fins. A total of 10 snorkels and 12 pairs of fins are rented. Renting a sno

rkel costs three times as much as renting a pair of fins. How much does it cost to rent a snorkel?

1 answer:

allsm [11]2 years ago

3 0

The snorkel costs $12 to rent!

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Ricky caught a snake that was 58.2 centimeters long. Which of these lengths has an 8 with a value of 1/10 the value of the 8 in

Anettt [7]

In every number, set every digit to zero that is not 8. Then look at the question again.

"Which of these lengths has 1/10 the value of 8.0?"

A. 800
B. 0.08
C. 80
D. 0.80

We hope the answer D is obvious to you.

... 0.80 = (1/10)×8.0

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The appropriate choice is ...

... D. 209.86

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2 years ago

- 9.9t = - 9.9 <br> solve for t

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T=1

Hope that helps bud!

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If an image of a triangle is congruent to the pre image then the scale factor of the dilation must be n=

Kazeer [188]

1. the image is multiplied by the scale factor. if the scale factor is 2, the image is twice the size of the preimage. if the scale factor is ⅓, the image is one third the size. only a scale factor of 1 preserves congruency

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2 years ago

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I would like to know the cost of c

Evgen [1.6K]

Answer:

1..75c

Step-by-step explanation:

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2 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago