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Oksanka [162]
3 years ago
6

Logan can paint a room in 12 hours. Marcus can paint the same room in 16 hours. How long does it take for both Logan and Marcus

to paint the room if they are working together?
Mathematics
1 answer:
Anettt [7]3 years ago
8 0
Hello,

The answer should be "4".

Reason:

Since Logan takes 12 hours and Marcus takes 16 i just subtract 16 and 12.

16-14=4

=4 hours

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
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Garry has 15 apples he gives 8 away how many apples does Garry have left?
Anuta_ua [19.1K]

Answer:

7 apples left

Step-by-step explanation:

So he has 15 apples. IF he gives aways 8, he only has 15-8= 7 apples left

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Select one of the factors of 3x2 10x 3
balu736 [363]
The polynomial is 3x^2 + 10x + 3 and the two factors are (3x+1) and (x+3), because

(3x+1)(x+3) = 3x^2 +9x + x +3 = 3x^2 + 10x +3
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Peter tracks the goals he scores in 9 soccer games. In two games Peter scored 2 goals each, in one game he scored 0 goals, in tw
Reika [66]

The average number of goals per game Peter scores, rounded to one decimal place is 1.6

<u><em>Explanation</em></u>

In first two games Peter scored 2 goals each. So, total score in two games = (2×2)= 4

In one game he scored 0 goal and in next two games he scored 3 goals each, so the total is (3×2)= 6

In the last four games he scored 1 goal in each, so the total is (1×4)= 4

So, the total score in all 9 games =4+0+6+4=14

Thus, the average number of goals = \frac{14}{9}= 1.5555....

So, the average number of goals per game Peter scores, rounded to one decimal place is 1.6

3 0
3 years ago
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1/4 plus 1/6 in simplest terms
Tasya [4]

Answer:

\frac{5}{12}

Step-by-step explanation:

\frac{1}{4}+\frac{1}{6}\\\\=\frac{3}{12}+\frac{2}{12}\\\\=\frac{5}{12}

7 0
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Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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