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3 years ago

Which pair shows equivalent expressions?

Mathematics

1 answer:

maria [59]3 years ago

7 0

The equivalent expression is A

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Do the i p t interest percentage and time the you add the percentage and interest then you multiply that by your time

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3 years ago

1)Give two numbers whose square roots add up to 5.

barxatty [35]

$\sqrt{9} +\sqrt{4}=5\\ \\\sqrt[3]{64}+\sqrt[3]{125}=9$

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4 0

4 years ago

Read 2 more answers

3 x 2 2/5 pleaseyou guys are supercalifragilisticexpialidocious

Yuliya22 [10]

7.2 is the answer i hope that helps!!!!!!!!!!!

7 0

3 years ago

SMART STUDENT PLEASE NEED GOOD MATH STUDENT

tangare [24]

Answer:

Therefore 'x' is equal to 65.4°

Step-by-step explanation:

In Right Angle Triangle ABC

∠ B = 90°

AC = Hypotenuse = 12

CB = Adjacent Side = 5

To Find:

∠ C = x

Solution:

In Right Angle Triangle ABC Cosine Identity we have

$\cos C = \frac{\textrm{side adjacent to angle C}}{Hypotenuse}\\$

Substituting the values we get

$\cos C = \dfrac{CB}{AC}=\dfrac{5}{12}$

$\angle C = \cos^{-1}(0.4166)=65.37=65.4\°$

Therefore 'x' is equal to 65.4°

8 0

3 years ago

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