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2 years ago

Can someone please help me

Mathematics

2 answers:

svetlana [45]2 years ago

8 0

The answer to this question is 56 your welcome

Nana76 [90]2 years ago

5 0

I think that it is 56

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H e l p (15 points :)

marishachu [46]

Answer:those are your classes and the last one is the classes you completed i think im not really sure tho but yeah, umm byeee

Step-by-step explanation:

3 0

2 years ago

4(4-w)=3(2w+2) find w

Travka [436]

Answer:

Step-by-step explanation:

4(4-w) = 3(2w+2)

Use distributive property:

16 - 4w = 6w + 6

+4w +4w

------------------------

16 = 10w + 6

-6 -6

-------------------------

10 = 10w

---- -----

10 10

1 = w

Hope this helped.

7 0

2 years ago

Read 2 more answers

PLS SOMEONE HELP ME

garik1379 [7]

Here are your answer's:

Number three would goes with "no relationship"...…
Number one would goes with "causation".....
Number two goes with "correlation"...…

Correlation is the relationship between more than two things which is why it would go with number to because literacy rates go with the increase of doctors. Causation is like a cause and effect so in this case the cause would be a increase of the population and the effect would be more people who don't have jobs. No relationship would go with number three since how snowfall doesn't have anything to do with personal income. (which is how much a person makes) 

Hope this helps!

7 0

3 years ago

If z = 3cis30°, z3 in rectangular form is _____ +_____ i.

oee [108]

The rectangular form is 3c/30=z
Z=10 3c//30
10z+70x + 17x-10z

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