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3 years ago

36 to the power -1/2

Mathematics

1 answer:

IRISSAK [1]3 years ago

6 0

0.166667
it was rounded

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A solution to the system y = 2x 7 and x +y= 5? Why or why not

andrezito [222]

Answer:

given that, y=2×7

y=14

and

x+y=5

x+14=5

x=5-14

x=9

x not equal to 5

5 0

2 years ago

Please help me! I will give brainly!

Viktor [21]

<h3>Answer: 0.34</h3>

============================================================

Explanation:

A = 0.61 is given to us

Pumpkin B travels 0.28 of a mile further than A does, so,

B = A+0.28 = 0.61+0.28 = 0.89

Pumpkin B travels 0.89 of a mile.

---------------------

Pumpkin C travels 0.06 of a mile more than B does, which means,

C = B + 0.06

C = 0.89 + 0.06

C = 0.95

Pumpkin C travels 0.95 of a mile.

---------------------

The last thing to do is to subtract the values of C and A

C - A = 0.95 - 0.61 = 0.34

Pumpkin C travels 0.34 of a mile further compared to pumpkin A.

---------------------

Side note: if you want to convert from miles to feet, then multiply by 5280.

For instance, 0.61 miles = 0.61*5280 = 3,220.8 feet.

5 0

3 years ago

Read 2 more answers

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

<u /> $\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$