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2 years ago

What is the percent increase from 1 to 6

Mathematics

1 answer:

Advocard [28]2 years ago

7 0

Answer: 20%

Step-by-step explanation:

1 5

100 x

100*5/1 = 20%

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Find the range of the graph

Alex

Answer:

4,-7 = 11

and

6, -9 = 15

3 0

2 years ago

Simply<br> (2k-9)^2+(k+2)^2=(2k+1)^2+(k-8)^2

vova2212 [387]

The square of a binomial expands as follows:

$(a+b)^2=a^2+2ab+b^2$

Expanding all the binomials, we have

$4k^2-36k+81+k^2+4k+4=4k^2+4k+1+k^2-16k+64$

Sum like terms:

$5k^2-32k+85=5k^2-12k+65$

Simplify terms appearing on both sides:

$-32k+85=-12k+65$

Move all terms involving k to the left and all numbers to the right:

$-20k=-20$

Divide both sides by -20:

$k=1$

6 0

3 years ago

Molly has 190 dollars in her bank account and spend 10 dollars every trip write an equation for molly

leonid [27]

10xt=190
T represents the amount of trips and 10 represents the amount of money she spends each trip. Since she has $190 already it will be on the other side of the equation after the equal sign

3 0

2 years ago

Obtain the sample standard deviation <br><br>9, 10, 11, 12 ,13 ,14 ,15 ,16 ,17

Marat540 [252]

Standard deviation (s): 2.7386127875258

8 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago