All categories

3 years ago

Plz show your work. Thx!! 9m=6

Mathematics

1 answer:

grandymaker [24]3 years ago

5 0

So first you would divide the 9 on both sides then 6 divided by 9 would be .6 repeating so m=.6 repeating.

You might be interested in

6) A plane is preparing to land. It is at an altitude of

slavikrds [6]

Answer:

Step-by-step explanation:

4800-1400 because 200*7 equals 1400.

3 0

3 years ago

An internet book company charges $7 for each paperback plus $2.75 for shipping and handling per order

Leto [7]

Step-by-step explanation:

is it addition I got 9.75

5 0

3 years ago

Carolina paid $15.80 for 4 hamburgers.

jarptica [38.1K]

Answer:

The cost of one hamburger is $3.95.

Step-by-step explanation:

15.80/4=3.95

6 0

2 years ago

Read 2 more answers

Multiply the following rational expressions and simplify the result

GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$ = $\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}$

= $\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}$

= $\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}$

= $\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}$

3 0

3 years ago

For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

3 0

3 years ago