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3 years ago

Plz i really need help!!!

Mathematics

1 answer:

stich3 [128]3 years ago

3 0

14. A
C=10+12h
I think one time fee is 10, the recurring fee is 12

15.B
c=2*12h=24h
initial value is going to be 0 because no one time fee, rate of change is 24

C=a*h+b(a is rate of change, and b is initial value)

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Answer these for me please pretty please

anzhelika [568]

Answer:

1. t = 48.654

2. $x = \frac{8}{3}$

3. $s = \frac{49}{4}$

4. $y = \frac{19}{12}$

Step-by-step explanation:

To solve equations using variables, perform inverse operations to undo each part and isolate the variable.

1. $\frac{t}{5.4} =9.01$

Multiply by 5.4 on both sides to undo division.

t = 9.01 * 5.4

t = 48.654

2. $\frac{3}{4}x = 2$

Multiply both sides by the reciprocal of 3/4 which is 4/3.

$x = 2*\frac{4}{3} \\x = \frac{8}{3}$

3. $s + \frac{1}{4} = 12\frac{1}{2}$

Convert 12 1/2 into an improper fraction.

$s + \frac{1}{4} = \frac{25}{2}$

Subtract 1/4 from both sides.

$s + \frac{1}{4} - \frac{1}{4} = \frac{25}{2} - \frac{1}{4}$

To subtract fractions without common denominators, convert 25/2 into a fraction with denominator of 4.

$s = \frac{50}{4} - \frac{1}{4}\\s = \frac{49}{4}$

4. $2\frac{2}{3} + y = 4 \frac{1}{4}$

Subtract 2 2/3 from both sides.

$2\frac{2}{3} - 2\frac{2}{3} + y = 4 \frac{1}{4}- 2\frac{2}{3}$

Convert each fraction into improper fractions and then to common denominators.

$y = \frac{17}{4} - \frac{8}{3}\\\\ y = \frac{51}{12} - \frac{32}{12} \\\\y = \frac{19}{12}$

6 0

4 years ago

12. Lin has $212 in her savings account. Her goal is to have $800 in the account in 21 weeks. How much does

uranmaximum [27]

Answer: Lin must save at least $28 a week

Step-by-step explanation:

I got my answer by subtracting the already earned $212 from the neccessary $800 and that got me $588. Then divided the $588 by the 21 weeks from her goal to get my answer of at leaast $28 a week.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%5Crm%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D%20%5

umka2103 [35]

Replace $x\mapsto \tan^{-1}(x)$ :

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx$

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace $x\mapsto\frac1x$ :

$\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx$

Then the original integral is equivalent to

$\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx$

Recall that for |x| < 1,

$\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}$

so that we can expand the integrand, then interchange the sum and integral to get

$\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx$

Integrate by parts, with

$u = \ln(x) \implies du = \dfrac{dx}x$

$du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

Recall the Fourier series we used in an earlier question [27217075]; if $f(x)=\left(x-\frac12\right)^2$ where 0 ≤ x ≤ 1 is a periodic function, then

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)$

Evaluate f and its Fourier expansion at x = 1/2 :

$\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)$

$\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$