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2 years ago

Mittens 85 Solve the proportion missing quantity. for the 5

Mathematics

1 answer:

Troyanec [42]2 years ago

6 0

What this doesn’t make sense

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Shen ran for president of the chess club, and he received 14 votes. There were 35 members in the club. What percentage of the cl

RUDIKE [14]

Answer:

40%

Step-by-step explanation:

14/35 members voted him. Lets convert this into a percentage:

14/35 * 100% = 14/35 *100/100 = 40%

I hope this helps! :)

3 0

2 years ago

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:

Fundamental Theorem of Calculus

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

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:

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

Integration by substitution

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

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

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

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

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

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

Substitute 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})$\end{minipage}}$

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

$\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:$

$\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}$

$\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:$

$\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Learn more about integration by substitution here:

brainly.com/question/28156101

brainly.com/question/28155016

4 0

1 year ago

Find the value of x.

sp2606 [1]

x = 64°

Step-by-step Explanation

x = 1/2[(360° - 2*58°)-2*58°]

x = 1/2[(360° - 2*58°) - 2*58°]

x = 1/2[(360° - 116°) - 116°]

x = 1/2[244° - 116°]

x = 1/2[128°]

x = 64°

8 0

3 years ago

The graph of the function f(x) = x3 – 7x – 6 intersects the x-axis at the points (–2, 0), (–1, 0), and (3, 0) as shown. Which ex

Alisiya [41]

The x-axis intercepts are the roots of the polynomial. So, the roots are x = - 2, x = - 1 and x = 3.

Therefore, the polynomial can be factored as:

(x - (-2)) * ( x - (-1) ) * (x - 3) = (x + 2)(x + 1)(x - 3).

Answer: (x + 2) (x + 1) (x - 3)

5 0

2 years ago

How much will $850 amount to be in three years if it is invested at 8% interest compounded quarterly for 3 years? A. $886.47 B.

Dimas [21]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\$850\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, so four times}
\end{array}\to &4\\

t=years\to &3
\end{cases}$

7 0

2 years ago

Mittens 85 Solve the proportion missing quantity. for the 5​

Mittens 85 Solve the proportion missing quantity. for the 5