Three painters can paint three walls in three minutes. how many painters are needed to paint 27 walls in nine minutes?

1 answer:

4 0

There is 3 variable, in this case, painters, walls and time(minutes). You need to find how many painters needed and provided information on the walls and time(minutes).

Increasing walls will cause more painter needed. It was reversed in time which was increased time will cause less painter needed.

In this case, you need to divide the variable walls with time since it reversed so: 27wall/9 minutes
It equal to : 3 walls/minute

Information provided is:
3 painters= 3 walls/3 minutes
Equal to : 3 paintes= 1 wall/minute

To make it 3 walls/minute, the painter needed will be:
(3 walls/minute) / (1 wall/minute) x 3 painters= 9 painters

Answer: 9 painters

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Find the perimeter of the pentagon MNPQR with vertices M(2, 4), N(5, 8), P(8, 4), Q(8, 1), and R(2, 1)

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Answer:

The pentagon MNPQR has a perimeter of 22 units.

Step-by-step explanation:

Geometrically speaking, the perimeter of the pentagon is the sum of the lengths of each side, that is:

$p = MN + NP + PQ + QR + RM$ (1)

$p = \sqrt{\overrightarrow{MN}\,\bullet \, \overrightarrow{MN}} + \sqrt{\overrightarrow{NP}\,\bullet \, \overrightarrow{NP}} + \sqrt{\overrightarrow{PQ}\,\bullet \, \overrightarrow{PQ}} + \sqrt{\overrightarrow{QR}\,\bullet \, \overrightarrow{QR}} + \sqrt{\overrightarrow{RM}\,\bullet \, \overrightarrow{RM}}$ (1b)

If we know that $M(x,y) = (2,4)$ , $N(x,y) = (5,8)$ , $P(x,y) = (8,4)$ , $Q(x,y) = (8,1)$ and $R(x,y) = (2,1)$ , then the perimeter of the pentagon MNPQR is:

$p =\sqrt{(5-2)^{2}+(8-4)^{2}} + \sqrt{(8-5)^{2}+(4-8)^{2}}+\sqrt{(8-8)^{2}+(1-4)^{2}}+\sqrt{(2-8)^{2}+(1-1)^{2}}+\sqrt{(2-2)^{2}+(4-1)^{2}}$ $p = \sqrt{3^{2}+4^{2}} + \sqrt{3^{2}+(-4)^{2}}+\sqrt{0^{2}+(-3)^{2}}+\sqrt{(-6)^{2}+0^{2}}+\sqrt{0^{2}+3^{2}}$