All Answers 356000 Answers

x^-8

use a dash

y=-3x+5 5x-4y=-3

'an answer'

If you mean 6 x 6, then no, 6 x 6 is not 12. 6+6=12, but not 6 x 6. 6 x 6 = 36.

25 is the answer

The symbol for square root is called a radical, and it looks like a check mark with a line that extends over the number you want to find the square root of.

The button on this calculator is 4 from the left and one up. It has the radical (Chack mark looking thing) and an "X" beside it.

Use this formula for your problem:

FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD. Where R=Interest rate per period, N=number of periods, P=periodic payment, FV=Future value.

FV=300{[ 1 + 0.06/12]^15 - 1 / 0.06/12}

FV=300{[1.005]^15 - 1 / 0.005}

FV=300 x 15.53655

FV=$4,660.96

Found the solution to the d) one so only a) and d) are solved for now

54yd Because Both sides Divided by 38= 54 yards

...44 :3

It ks solve able i took alot of time n only half throuh i got 4.700291133*10^11

Its 8

\(2xy-6xz\)

If you're searching for a simplified expression, that would be \(3x+2.\)

log5 1/3125=x

x=Log(1/3125) / Log(5)

x= -5

in caculator terms it would be 5e-7 but in actual human terms it would be like this 10 divided by itself 7 times because your subtracting an exponet therefore it would be 0.000001 now if you multiply the equation it would be 5 times 0.000001 wich is just 0.000005

The reciprocal of 7 and ALL the multiples of 7 will repeat sooner or later because they are all rational numbers. Even your 1/49 repeats every 42 digits. You may not see it on your calculator because the display is limited on most calculators to 10 to 20 digits. So, 1/49 =

0.020408163265306122448979591836734693877551020408163265306.................etc.

a negative minus a positive gives you a negative 

14 ?

you can just say the awnser is undefinable 

integrating: 

\(\int dx ~ 2x \sin^{3}(x) \)

1. Let

\(\begin{array}{|rcll|} \hline \int dx ~ \sin^3(x) &=& \int dx ~ \sin^2(x)\cdot \sin(x) \\ &=& \int dx ~ [ 1-\cos^2(x) ]\cdot \sin(x) \\ && u = \cos(x) \\ && du = -\sin(x) dx \\ &=& \int du ~ \frac{1}{-\sin(x)} \cdot (1-u^2) \cdot \sin(x) \\ &=& -\int du ~ (1-u^2) \\ &=& -\int du +\int du ~ u^2 \\ &=& -u + \frac13 \cdot u^3 \\ &=& \mathbf{-\cos(x) + \frac13 \cdot \cos^3(x)} \\ \hline \end{array} \)

2. Let

\(\begin{array}{|rcll|} \hline \int dx ~ \cos^3(x) &=& \int dx ~ \cos^2(x)\cdot \cos(x) \\ &=& \int dx ~ [ 1-\sin^2(x) ]\cdot \cos(x) \\ && u = \sin(x) \\ && du = \cos(x) dx \\ &=& \int du ~ \frac{1}{\cos(x)} \cdot (1-u^2) \cdot \cos(x) \\ &=& \int du ~ (1-u^2) \\ &=& \int du - \int du ~ u^2 \\ &=& u - \frac13 \cdot u^3 \\ &=& \mathbf{\sin(x) - \frac13 \cdot \sin^3(x)} \\ \hline \end{array}\)

3. Let

\(\begin{array}{|rcll|} \hline \int dx ~ x \cdot \sin^3(x) \\ && u = x \\ && u' = 1 \\ && v = -\cos(x) + \frac13 \cdot \cos^3(x) \\ && v' = \sin^3(x) \\ &=& x \cdot [-\cos(x) + \frac13 \cdot \cos^3(x)] - \int dx ~ 1\cdot [-\cos(x) + \frac13 \cdot \cos^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \int dx ~ \cos(x) - \int dx ~ \frac13 \cdot \cos^3(x) \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot \int dx ~ \cos^3(x) \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot [ \sin(x) - \frac13 \cdot \sin^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot \sin(x) + \frac19 \cdot \sin^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \frac23 \cdot \sin(x) + \frac19 \cdot \sin^3(x) \\\\ \int dx ~ 2\cdot x \cdot \sin^3(x) &=& -2x \cdot \cos(x) + \frac23 \cdot x \cdot \cos^3(x) + \frac43 \cdot \sin(x) + \frac29 \cdot \sin^3(x) \\ &=&\mathbf{\frac{1}{9} \cdot [ -18x \cdot \cos(x) + 6 x \cdot \cos^3(x) + 12 \cdot \sin(x) + 2 \cdot \sin^3(x) ] }\\ \hline \end{array} \)

4. Let

\(\begin{array}{|rcll|} \hline && \cos^3(x) = \frac14 \cdot [3\cdot \cos(x) + \cos(3x) ] \\ && \sin^3(x) = \frac14 \cdot [3\cdot \sin(x) - \sin(3x) ] \\ \hline \end{array}\)

5. Let

\(\begin{array}{|rcll|} \hline && \int dx ~ 2\cdot x \cdot \sin^3(x) \\ &=& \frac{1}{9} \cdot [ -18x \cdot \cos(x) + 6 x \cdot \cos^3(x) + 12 \cdot \sin(x) + 2 \cdot \sin^3(x) ] \\ &=& \frac{1}{9} \cdot \{ -18x \cdot \cos(x) + 6 x \cdot \frac14 \cdot [3\cdot \cos(x) + \cos(3x) ] + 12 \cdot \sin(x) + 2 \cdot \frac14 \cdot [3\cdot \sin(x) - \sin(3x) ] \} \\ &=& -2x \cdot \cos(x) + \frac16 \cdot x\cdot [3\cdot \cos(x) + \cos(3x) ] + \frac{4}{3} \cdot \sin(x) + \frac{1}{18} \cdot [3\cdot \sin(x) - \sin(3x) ] \\ &=& -2x \cdot \cos(x) + \frac12 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac{4}{3} \cdot \sin(x) + \frac{1}{6} \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) \\ &=& \mathbf{ - \frac32 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac32 \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) }\\ \hline \end{array} \)

or

\(\begin{array}{|rcll|} \hline && \int dx ~ 2\cdot x \cdot \sin^3(x) \\ &=& - \frac32 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac32 \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) \\ &=& \mathbf{ \frac{1}{18} [ - 27 \cdot x\cdot \cos(x) + 3 \cdot x\cdot \cos(3x) + 27 \cdot \sin(x) - \sin(3x) ] } \\ \hline \end{array}\)

2/3 = 8/12

2/4 = 6/12

8 - 6 = 2

Therefore, 2/3 take away 2/4 is equal to 2/12, which can be simplified to 1/6.

Close. Let's see where you went wrong though.

-3u=15

divide by -3

-3u=15

 -3    -3

u=-5

So the answer is -5. Good try though!

what is 15 scientific notainon

-3u = +15

Find the value of u.

The value of u is -5. A negative multiplied by a negative is a positive. If the answer was -15, then the value of u would be +5, since a negative multiplied by a positive (vice versa) equals to a negative.

Hopefully that helps.