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padilas [110]
3 years ago
10

Chin correctly translated the following phrase into an algebraic expression. “One-fifth less than the product of seven and a num

ber” which represents chin’s phrase?
Mathematics
2 answers:
Harrizon [31]3 years ago
6 0

<em>Answer:</em>

<h2>D is the answer on Edgen.</h2>

<em>Step-by-step explanation:</em>

7n - 1/5

kirill [66]3 years ago
3 0

Let's assume the number be x.

So, product of seven and a number can be written as 7*x.

Now the given statement is “One-fifth less than the product of seven and a number” which means we need to take out 1/5 from 7x. Hence, it can be converted into an algebraic expression as follows:

7x-\frac{1}{5}

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$20 because 2x4=8 3x4=12 12+8=20

Step-by-step explanation:

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3 years ago
What is (4a²)³÷8a²? can you please answer in normal number form (not like 3²)
AleksandrR [38]

Answer:

8a^4

Step-by-step explanation:

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Now we've got\frac{64a^6}{8a^2}=8a^4.

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If the company is planning to produce 90,000 containers of crunchy peanut butter
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Is there more to the problem? 
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3 years ago
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
please help me with these two problems PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE ​
lisov135 [29]
You can use Math-way for problems like this !! ... it’s an app it’s good for giving you answers on math equations
4 0
3 years ago
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