Somebody helppppp pleassse

Mathematics

1 answer:

Lapatulllka [165]4 years ago

4 0

The regular exagon (polygon in the figure) is equilateral and equiangular.

You can divide all the sides in triangles...

That thing that measures 5√3 cm is the apotheme.

The apotheme divides the equilateral triangles in half so the halves are 30° - 60° - 90° triangles and then we know that the apotheme is x√3 so...

x√3 = 5√3 -> divide both for √3

x = 5

This is the base of the halves, thus the base of the entire triangle is 10. We find the perimeter of the exagon -> 10*6 = 60 cm

We can find the area = 1/2 * perimeter * apotheme

1/2*60*5√3 = 150√3 cm²

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Q20-23 with steps pls

just olya [345]

20: Let $x$ be the amount of money. If this amount is shared between 8 people, each person will get x/8 dollars. But there actually are 6 people, so everyone is getting x/6 dollars. We know that this difference results in 30 dollars more, so we have

$\dfrac{x}{6}=\dfrac{x}{8}+30 \iff \dfrac{x}{6}-\dfrac{x}{8}=30$

Rearrange the left hand side as

$\dfrac{4x-3x}{24}=30$

And multiply both sides by 24 to get

$x=720$

21: Let $M$ and $J$ be the number of marbles owned by Mike and Judy, respectively. At the beginning, we have

$M=3J+11$

If they both get 9 more marbles, Mike will have $M+9$ marbles, and Judy will have $J+9$ marbles. So, we have

$M+J+18=93 \iff M+J=75 \iff M=75-J$

Plug this value in the first equation and we have

$75-J=3J+11 \iff 4J=64 \iff J=16$

And we deduce

$M=16\cdot 3 + 11=59$

So, the difference is

$M-J=59-16=43$

22: Let $x,y,z$ be the number of $10, $20 and $50 coupon, respectively. We're given:

$\begin{cases}x=2y+3\\z=\frac{1}{2}y\\x+y+z=38\end{cases}
We can write the third equation by substituting the expressions for x and z:
[tex]x+y+z=(2y+3)+y+\left(\dfrac{1}{2}y\right)=38$