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3 years ago

Simplify. Show your work.

Mathematics

1 answer:

alex41 [277]3 years ago

8 0

1 5/6 is the correct answer.

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Please help am stuck ASAP

Aloiza [94]

Answer:

linear

Step-by-step explanation:

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1 year ago

One third divided by eight

stellarik [79]

Answer:

1/24

Step-by-step explanation:

1/3÷ 8

Copy dot flip

1/3 * 1/8

1/24

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3 years ago

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A container holds 6 quarts of lemonade. How much is this in cups?

sdas [7]

Well, 4 cups = 1 quart, so 4 *6= 24 cups in 6 quarts of lemonade.

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3 years ago

PLEASE HELP!!!!!!!!!!! Kim wants to know if the number of words on a page in her math book is generally more than the number of

omeli [17]

It's D, because the mean doesn't matter as much if it's only taken in a small variability. The mean was selected from 25 random pages in the book, that is why it is right.

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3 years ago

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Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago