All categories

2 years ago

Question please answer from 3 to 9 do not say anything if you don't know the answer

Mathematics

1 answer:

vovikov84 [41]2 years ago

6 0

Answer:

1) 4 pounds 2) 5.25 kilograms 3) 1.5 gallons 4) 1.33 miles 5) 4.5 feet tall 6) 3 cups 7) 4.5 inches 8)1.166 9) 2.1 boxes

Step-by-step explanation:

:P ur welcome

You might be interested in

Footlocker buys sneakers for $60 each and marks up the price by 30% so they can make a profit. What is the selling price of the

pantera1 [17]

Answer:

now it would be $78

4 0

2 years ago

An item is regularly priced at$90. Henry bought it on sale for 70% off the regular price. How much did Henry pay?

Neko [114]

70 times .9 is 63 so 90 minus 63 is27 so it was $27

3 0

2 years ago

Read 2 more answers

Bryan wants to convert a speed of 85 kilometers per hour into the speed in meters per second what is the correct conversion

lara31 [8.8K]

Answer:

23.6111

hehe

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

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

Select the reciprocal of 6/5?

ale4655 [162]

The answer is 5/6 because you need to flip the numbers.

8 0

3 years ago

Question please answer from 3 to 9 do not say anything if you don't know the answer​

Question please answer from 3 to 9 do not say anything if you don't know the answer