All categories

3 years ago

WILL MARK BRAINLIST 10 to the power of 8 =

Mathematics

2 answers:

Nataly_w [17]3 years ago

6 0

Answer:

10^8= 100,000,000

Step-by-step explanation:

10×10×10×10×10×10×10×10= 100,000,000

myrzilka [38]3 years ago

6 0

10^8=100000000 hopes this helps

You might be interested in

Which relation is also a function?

slavikrds [6]

Answer:

Hi san, A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y. That is, given an element x in X, there is only one element in Y that x is related to. For example, consider the following sets X and Y.

This should help you a bit.

6 0

3 years ago

How to graph the line y=4/3x

ankoles [38]

Answer:

make a table of values

Step-by-step explanation:

then plot using those values

3 0

3 years ago

Convert 131 57’ 32” to decimal degree

Arlecino [84]

Answer:

131.95888888888888

Step-by-step explanation:

Your Welcome :)

5 0

3 years ago

A car moving at an initial velocity of 20 meters/second accelerates at the rate of 1.5 meters/second2 for 4 seconds. The car’s f

sineoko [7]

The initial velocity: v o = 20 m/s.And : a = 1.5 m / s² , t = 5 s.The final velocity formula:v = v o + a · tv = 20 + 1.5 · 5v = 20 + 7.5v = 27.5 m /sAnswer: The car`s final velocity is 27.5 meters / second.

4 0

2 years ago

Read 2 more answers

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago