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2 years ago

Complete the table and then graph the function.

Mathematics

2 answers:

nasty-shy [4]2 years ago

6 0

Can we submit by paper??????????????

Gekata [30.6K]2 years ago

4 0

Answer:

You plug in x and solve for each value. Look at attachment for answers

Step-by-step explanation:

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Please help been stuck on this for 40 minutes

choli [55]

we have given y interms of x.

we have given that add 2 to x.

which means $y=x+2$

so we have $x=2 ,y=x+2=2+2=4$ .

so when x $=2 ,y=4$ .

$x=4,y=x+4=4+2=6.$ .

$y=x+2,y=6+2=8$ .

3 0

3 years ago

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A survey has a margin of error of +/– 3%. What is the range of number of people who will vote for a candidate if 87 of the 110 p

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

I just answered it

6 0

3 years ago

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the centers of Marathon County Bayfield County and Dodge County all lie along a straight line if Marathon County is equidistant

Evgen [1.6K]

Marathon C is in the middle.

From Marathon C to Dodge C is 487

Then From Marathon C to Bayfield C is also 487

And from Dodge C to Bayfield C is 487 + 487 = 974 miles

5 0

3 years ago

Need help please!!!!!!!

stepan [7]

Step-by-step explanation:m okay so the answer is 56 x= 56

6 0

2 years ago

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