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3 years ago

why is it important to write any zero in the correct place value proposition when using the break apart strategy to add?

Mathematics

1 answer:

Katarina [22]3 years ago

7 0

If u don't it could change the value of the number

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Which of the following shows the solution to m+5 >-1

Ne4ueva [31]

The first one m > -6

6 0

2 years ago

Read 2 more answers

Help pls

timurjin [86]

Answer:

s> 144 2/3

Step-by-step explanation:

I took the test in k12 and i got it right so yea !

3 0

1 year ago

If 2x/3 = 8/4, then x=

velikii [3]

Answer:

first cross multiply

2x×4=3×8

8x=24

then, divide both sides by "8" so x could be alone

x=24/8

x=3

i hope this helps, please give me brainliest

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Step-by-step explanation:

3 0

2 years ago

What are the steps she used ?

Serjik [45]

1. Add x on both sides of the equation.
2. Subtract 5 from both sides of the equation.
3. Divide both sides of the equation by 3.

If these aren’t options, tell me the options for each dropdown in the comments and I can help further.

Please consider marking this answer as Brainliest to help me advance.

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