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2 years ago

Can you please please help

Mathematics

2 answers:

dmitriy555 [2]2 years ago

8 0

He ate 1/8 of the pizza

Oxana [17]2 years ago

5 0

120 then 120 times 5= 600 then 600 times 9= 5400 then 5400 times 3= 16200 times 5= 81000 then 8100 times 5= 85000

85000 is your answer.

Don't forget to show your work!!

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Suppose that the water level of a river is 34 feet and that it is receding at a rate of her foot per day right in equation and h

Ghella [55]

Answer: it will take 9 days

Step-by-step explanation:

The water level was initially at 34feet it was receding at the rate of a foot per day. The rate at which it was receding is linear, thus, it is in an arithmetic progression. The formula for the nth term of an arithmetic progression is expressed as

Tn = a+(n-1)d

Where

a is the first term if the sequence

d is the common difference

n is the number of terms.

From the information given,

a = 34 feet because it is the initial height

n is the number of days it will take to get to 26 ft

d = -1 because it is decreasing by 1 foot per day.

Tn = T26 = 26 feet. Therefore, the equation will be

Tn = 34 -1(n-1)

To find for T26,

26= 34 + (n - 1)-1

26 - 34 = -n + 1

n - 1 = 8

n = 8 + 1 = 9

8 0

3 years ago

How do you solve this?

Alchen [17]

The answer is (3x^2 - x +2 )= to AC
Area = 1/2 |BD| |AC|
(9x^3 - 3x^2 + 6x) = 1/2 (AC) 6x
3x ( 3x^2 - x +2 )= 1/2 (AC) 2(3x)
3^2 - x +2 = AC

3 0

3 years ago

The gcf of 42 and 70 is

Charra [1.4K]

The gcf of 42 and 70 is 14

7 0

2 years ago

Read 2 more answers

128.5 divided by 4.5

Ganezh [65]

Answer:

your answer is 28.5

7 0

3 years ago

Read 2 more answers

Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$

Then add all these boxed integrated together then we'll get:

$\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}$

Expand 4 in the expression:

$\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}$

Therefore, the answer is:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

4 0

8 months ago