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2 years ago

What is 935 divided by 9?

Mathematics

2 answers:

Dimas [21]2 years ago

7 0

103 with a remainder of 8

Tema [17]2 years ago

7 0

Answer:

103.889

Step-by-step explanation:

I hope this helps :)

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Aubrey invested $61,000 in an account paying an interest rate of 1.9% compounded continuously. Assuming no deposits or withdrawa

Hitman42 [59]

Answer: 9.9 years.

Step-by-step explanation:

If interest is compounded continuously, then formula to compute final amount A = $Pe^{rt}$ , where P =initial amount, r= rate of interest , t=time.

Given: P= $61,000, r= 1.9% =0.019 , A = $ 73600

Substitute all values in formula

$73600=61000e^{0.019t}\\\\\Rightarrow\ \dfrac{73600}{61000}=e^{0.019t}\\\\\Rightarrow\ 1.20655738=e^{0.019t}$

Taking natural log on both sides

$\ln (1.20655738)=0.019t\\\\\\ 0.18777116=0.019t\\\\\\ t=\dfrac{0.18777116}{0.019}\\\\\\ t=9.88269263\approx9.9\ \text{years}$

Hence, the required time = 9.9 years.

4 0

3 years ago

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

Lynna [10]

Answer:

would multiply everything by the common denominator of 4

-3/4m - 1/2 = 2 + 1/4m....multiply everything by 4 and u get :

-3m - 2 = 8 + m

that will get rid of the fractions....but u could multiply by -4 ...giving u :

3m + 2 = -8 - m

Step-by-step explanation:

7 0

2 years ago

1 . If x = 3+2√2, then find the value of x- 1-x

Bess [88]

Step-by-step explanation:

Thanks don't forget to follow up...

5 0

2 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

PLEASE HELP! ! ! urgent!

sertanlavr [38]

Answer:

$6(\frac{1}{3} \times\frac{7}{5} )=(6\times\frac{1}{3} )\frac{7}{5}$

$Answer: B$

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Hope it helps...

Have a great day!!

8 0

3 years ago