2m^2n^4/6m^5n^3 assume no variable equals zero.

Which Answer is closet to the difference?

bearhunter [10]

Well the difference is -43 so I'm pretty sure it's gonna be A

5 0

3 years ago

Sarah has a collection of nickels, dimes, and quarters worth $28.75. She has 10 more dimes than nickels and twice as many quarte

Ymorist [56]

Answer:

Step-by-step explanation:

Write an equation for each statement:

:

"Sarah has a collection of nickels, dimes, and quarters worth $15.75."

.05n + .1d + .25q = 15.75

:

"She has 10 more dimes than nickels"

d = n + 10

:

"twice as many quarters as dimes."

q = 2d

:

How many coins of each kind does she have?

:

Take the 2nd equation and arrange it so n is in terms of d also

n + 10 = d

n = (d - 10)

:

In the 1st equation substitute (d-10) for n and 2d for q:

.05n + .1d + .25q = 15.75

:

.05(d-10) + .1d + .25(2d) = 15.75

:

.05d - .5 + .1d + .5d = 15.75

:

.65d - .5 = 15.75

:

.65d = 15.75 + .5

:

.65d = 16.25

:

d = 16.25/.65

:

d = 25 dimes

:

Remember the statement "twice as many quarters as dimes."

q = 2(25)

q = 50 quarters

:

The statement "She has 10 more dimes than nickels"

n = 25 - 10

n - 15 nickels

:

Check our solutions

.05(15) + .1(25) + .25(50) =

.75 + 2.50 + 12.50 = 15.75 proves our solutions

8 0

2 years ago

Simplify -2[-6(-4 7)] 36 -24 -36

7nadin3 [17]

-2(-53)×36-24-36 》3816-60=3756

3 0

4 years ago

Read 2 more answers

What is -5/4 to the 2nd power?

Vsevolod [243]

Answer:

$\frac{25}{16}$

Step-by-step explanation:

$(-\frac{5}{4})^2\\\\ \text {Apply power of a fraction rule: } (\frac{a}{b})^x=\frac{a^x}{b^x}\\\\(-\frac{5}{4})^2 = \frac{-5^2}{4^2}=\frac{25}{16}\\\\\boxed{(-\frac{5}{4})^2=\frac{25}{16}}$

8 0

3 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

3 years ago