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3 years ago

Simplify 150 -(0.4 x 150)+15

Mathematics

1 answer:

Rufina [12.5K]3 years ago

5 0

150-(0.4 x 150)+15= 105

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A box contains 5 blue marbles 3 red marbles and 6 white marbles. if three marbles are drawn, without replacement, what is the pr

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The probability that all three are blue it's 3:14

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3 years ago

Solve for c: c - 12 = 3

vlabodo [156]

Steps to solve:

c - 12 = 3

~Add 12 to both sides

c = 15

Best of Luck!

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3 years ago

Read 2 more answers

Isaac is organizing a 5-kilometer road race. The safety committee

Artyom0805 [142]

Answer:

<h2>10 volunteers are more than recommended</h2>

Step-by-step explanation:

The recommended number of volunteers is five (5)

Since the the distance of the race is 5km,

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Hence ten (10) volunteers is more than enough

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3 years ago

Ten boys and twelve girls decide to rent a 16-passenger van and a 6-passenger to drive to a rock concert in a near by city. If t

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A) The probability that there are no boys in the car is given by:
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3 years ago

Find the area of the shaded region

o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

$\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)$

$A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}$

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2 years ago