Ask question

Ask question

All categories

2 years ago

7

it takes 20 pounds of seed to completely plant a 3-acre field. How many pounds of seed are needed per acre?

1 answer:

lianna [129]2 years ago

4 0

Answer:

it will take 6.666667 acres

Step-by-step explanation:

so you do 20/3 and you need to find the unit rate so

20/3 = 6.6667/1

hopefully this helps

You might be interested in

1. Here is a rule to make a list of numbers: Each number is the sum of the

boyakko [2]

Answer:

what the rest of the problem?

Step-by-step explanation:

8 0

2 years ago

Find 8 and one third percent of 144

madam [21]

$\bf 8\frac{1}{3}\implies \cfrac{8\cdot 3+1}{3}\implies \cfrac{25}{3}
\\\\\\
now\qquad \cfrac{\frac{25}{3}}{100}\implies \cfrac{\frac{25}{3}}{\frac{100}{1}}\implies \cfrac{25}{3}\cdot \cfrac{1}{100}\implies \cfrac{1}{3\cdot 4}\implies \cfrac{1}{12}
\\\\\\
8\frac{1}{3}\%\ of\ 144\implies \left( \cfrac{\frac{25}{3}}{100} \right)\cdot 144\implies \cfrac{1}{12}\cdot 144\implies \cfrac{144}{12}\implies 12$

3 0

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

Work out size of angle x

myrzilka [38]

Answer:

Step-by-step explanation:

Comment

You are given values for the adjacent side of <x and the hypotenuse of a right triangle. This defines the Cosine of an angle

Equation

Cos(x) = a/c the way you have labeled it.

Givens

a = 9

c = 18

Solution

Cos(x) = a / c

Cos(x) = 9 / 18

cos(x) = 1/2

x = cos-1(1/2)

x = 60 degrees

Answer

x = 60

7 0

2 years ago

Read 2 more answers

A shoppers discount club charges a monthly fee of $15 and sells gasoline for $2.05 per gallon. The gas station across the street

svet-max [94.6K]

$15.75 dnnddbbd d d d Dbdjdhhdhdvdg

3 0

2 years ago