All categories

3 years ago

the ratio of white pens to green pens is 2 to 8. how many green pens are there if there are 18 white pens?

1 answer:

5 0

There are 72 green pens.

There is 9 times as many white pens from 2 to 18, so there has to be 9 times as many green pens because of the ratio. 9 times 8 is 72.

You might be interested in

Given an equation y= -8x -4. Identify the slope and y-intercept

Naddika [18.5K]

Answer:

Slope = - 8

y-intercept = - 4

Step-by-step explanation:

$y= -8x -4 \\ \\ equating \: it \: with \\ \\ y = mx + b \\ \\ slope \: (m) = - 8 \\ \\ y - intercept \: (b) = - 4$

8 0

2 years ago

Which best describe the meaning of this function notation ? Cost(weight)

labwork [276]

It tells you how temperature varies with pressure :)

3 0

3 years ago

Read 2 more answers

A half-circle is joined to an equilateral triangle with side lengths of 12 units. What is the perimeter of the resulting shape?

krek1111 [17]

Answer:

P = 42.84 cm

Step-by-step explanation:

A half-circle is joined to an equilateral triangle with side lengths of 12 units.

We need to find the perimeter of the resulting shape.

The perimeter of the half cirle is πr.

The diameter of the circle is 12. It means radius is 6 units.

An equilateral triangle has 3 equal sides. So, perimeter of the triangle is given by :

P = πr + 12 +12

= 3.14(6) + 24

= 42.84 cm

Hence, the perimeter of the resulting shape is 42.84 cm.

8 0

3 years ago

The legs of a right triangle are 3 units and 6 units. What is the length of the hypotenuse? Round your answer to the nearest hun

garri49 [273]

C = $\sqrt{3 ^{2} + 6 ^{2} }$ = 6.7082

Round 6.7082
6.71

6.71 is your answer

8 0

3 years ago

Read 2 more answers

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.

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

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

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

5 0

2 years ago