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3 years ago

Simplify the following expression. 0.82 / 0.2

2 answers:

8 0

4.1 when you solve out between the numbers of .82 and .2 it equals to 4.1 you can also solve it on a calculator it makes it pretty easy

AlekseyPX3 years ago

5 0

Answer:

0.41

Step-by-step explanation:

0.82 = 0.2 x 0.41

0.82 / 0.2

= ( 0.2 x 0.41 ) / 0.2

= 0.41

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PLZ ANSWER !!!! BRAINLIEST!!!

elena-14-01-66 [18.8K]

Answer: 238,400,000 is predicted to wear corrective lenses.

Step-by-step explanation:

We can use a ratio table to solve this problem.

So far, with the data given, we can construct a ratio table that looks like this:

Population: 1000, 320,000,000

Cor. Lenses: 745, ?

We know 745 people out of 1000 wear corrective lenses, yet how many out of 320 million people wear them?

To solve this, we need to figure out how many times 1000 is away from 320 million. We can find this out by dividing 320,000,000 by 1000. This results in 320,000. This meant that 1000 was multiplied 320,000 times to get to 320 million. However, what we do to one side of the ratio table, we must do it to the other.

This means we have to multiply 745 by 320,000. This results in 238,400,000. Out of 320 million Americans, 238.4 million wear corrective lenses.

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4 years ago

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Pls answer I give brainliest thank you! Number 4

soldi70 [24.7K]

Answer:

Step-by-step explanation:

I did the work but like where did that cube go lol

4 0

3 years ago

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PLEASE HELP QUICKLY

NikAS [45]

K=4 I think is the answer.

3 0

3 years ago

Can you just answer this question?

harkovskaia [24]

It should be B 7 Pack for $1.82

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3 years ago

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Use the form of the definition of the integral given in the theorem to evaluate the integral. ∫ 0 − 2 ( 7 x 2 + 7 x ) d x

Murrr4er [49]

Answer:

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

Step-by-step explanation:

The definite integral of a continuous function <em>f</em> over the interval [a,b] denoted by $\int\limits^b_a {f(x)} \, dx$ , is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,

$\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty} \sum_{i=1}^{n}\Delta x \cdot f(x_i)$

where $\Delta x = \frac{b-a}{n}$ and $x_i=a+\Delta x\cdot i$

To evaluate the integral

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx$

you must:

Find $\Delta x$

$\Delta x = \frac{b-a}{n}=\frac{0+2}{n}=\frac{2}{n}$

Find $x_i$

$x_i=a+\Delta x\cdot i\\x_i=-2+\frac{2i}{n}$

Therefore,

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} f(-2+\frac{2i}{n})$

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx=\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})$

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})$ $\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} 4-\frac{8i}{n}+\frac{4i^2}{n^2} -2+\frac{2i}{n}\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2$

$\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]$

We can use the facts that

$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}$

$\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(\frac{n^2-3n+2}{n^2}\right)\\\\\mathrm{Divide\:by\:highest\:denominator\:power:}\:1-\frac{3}{n}+\frac{2}{n^2}\\\\\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)\\\\\frac{14}{3}\left(\lim _{n\to \infty \:}\left(1\right)-\lim _{n\to \infty \:}\left(\frac{3}{n}\right)+\lim _{n\to \infty \:}\left(\frac{2}{n^2}\right)\right)\\\\\frac{14}{3}\left(1-0+0\right)\\\\\frac{14}{3}$

Thus,

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

5 0

4 years ago