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3 years ago

0.00065 in Standard form

Mathematics

2 answers:

Rus_ich [418]3 years ago

4 0

I think you meant scientific form well the answer is 6.5 x 10^-4

umka2103 [35]3 years ago

3 0

Try 6.5x10^4
Pretty sure that’s it, If not I’m sorry!

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Carissa and Louann both have savings accounts. Carissa has $250 in her account and deposits $80 per month. Louann has $1230 in h

zlopas [31]

Carissa value after x months=current+amount deposited
louan value after x months=current-amount taken out

carissa depositied=amount per month times x months=80x
louan take out=amount per month tiems x months=60x

when wil amount be equal
se equal

cariss=lousa
250+80x=1230-60x
add 60x both sides
250+140x=1230
minus 250 both sides
140x=980
divide both sides by 140
x=7

find how much that is
250+80(7)=250+560=810

7 months both have $810

8 0

3 years ago

WILL GIVE BRAINLISET!!! Drag and drop the answers into the boxes to identify the vertex and axis of symmetry of this function.

tensa zangetsu [6.8K]

Answer:

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Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I need to find the distance!whats the distance?

posledela

Answer:

Exact Form:

√ 61 (square root )

Decimal Form:

7.81024967 …

Step-by-step explanation:

Use the distance formula to determine the distance between two points.

6 0

2 years ago

To save money, you put $200 in your bank account each week. After saving for 4 weeks, you have $1,700 dollars in your account. W

kati45 [8]

it is c y−1,700=200(x−4)

8 0

3 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