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Ostrovityanka [42]
3 years ago
8

Help please! Will pick the Brainliest. (sorry if it’s sideways)

Mathematics
1 answer:
KatRina [158]3 years ago
3 0

1.

(x^2y^8)^\frac{2}{3}=x^{2\cdot\frac{2}{3}}y^{8\cdot\frac{2}{3}}=x^{\frac{4}{3}}y^{\frac{16}{3}}=x^{1\frac{1}{3}}y^{5\frac{1}{3}}=x^1x^\frac{1}{3}y^5y^\frac{1}{3}=xy^5\sqrt[3]{x}\sqrt[3]{y}=\boxed{xy^5\sqrt[3]{xy}}

Answer B)

2.

\frac{1}{3}\left(\frac{2}{15}x-\frac{2}{3}\right)>x+\frac{1}{5}\quad|\cdot3\\\\\\\frac{2}{15}x-\frac{2}{3}>3x+\frac{3}{5}\quad|\cdot15\\\\\\2x-\frac{30}{3}>45x+\frac{45}{5}\\\\2x-10>45x+9\\\\-10-9>45x-2x\\\\-19>43x\quad|:43\\\\\boxed{x

Answer B)

3.

\sqrt{14xy}\cdot\sqrt{12}\cdot\sqrt{30y}=\sqrt{14\cdot12\cdot30}\cdot\sqrt{xy^2}=\sqrt{2\cdot7\cdot3\cdot4\cdot2\cdot3\cdot5}\cdot\sqrt{xy^2}=\\\\=\sqrt{2^2\cdot3^2\cdot2^2\cdot5\cdot7}\cdot\sqrt{xy^2}=2\cdot3\cdot2\cdot\sqrt{35}\cdot y\cdot\sqrt{x}=\boxed{12y\sqrt{35x}}

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50.7202

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If y varies directly with x, and y = 10 when x = 4, write the direct linear variation equation
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The town of Madison has a population of 25{,}00025,00025, comma, 000. The population is increasing by a factor of 1.121.121, poi
cestrela7 [59]

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Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

P(t)=a(b)^t

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3 0
3 years ago
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Oksana_A [137]

Answer:

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Please make me the brainliest because I havent got even 1 yet :(



5 0
3 years ago
Find the period of the function. y=3 sin x/8
Alexus [3.1K]

Answer:

The period of given function is  Period = 16\pi

So, Option B is correct.

Step-by-step explanation:

In this question we need to find the period of the function y= 3 sin x/8

The formula used to find period of function is: \frac{2\pi }{b}

We need to know the value of b.

To find the value of b we compare the standard equation with the equation of function given.

Standard Equation: y = a sin(bx - c) +d

Given Equation: y= 3 sin(x/8)

Comparing we get:

a= 3

b= 1/8

c= 0

d=0

So, we get the value of b i.e 1/8. Putting it in the formula to find period of given function.

Period = \frac{2\pi }{b}

Period = \frac{2\pi }{\frac{1}{8}}

Solving,

Period = 2\pi *8

Period = 16\pi

So, the period of given function is  Period = 16\pi

5 0
3 years ago
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