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adelina 88 [10]
3 years ago
8

I need help on this question ASAP tysm

Mathematics
1 answer:
Ray Of Light [21]3 years ago
5 0
I am stuck on one do those as well, try Socratic
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The side of each of the equilateral triangles in the figure is twice the side of the central regular hexagon. What fraction of t
lana [24]

Answer:

The fraction is 1/4

Step-by-step explanation:

we know that

The area of an equilateral triangle, using the law of sines is equal to

A=\frac{1}{2}x^{2}sin(60^o)

A=\frac{1}{2}x^{2}(\frac{\sqrt{3}}{2})

A=x^{2}\frac{\sqrt{3}}{4}

where

x is the length side of the triangle

In this problem

Let

b ----> the length side of the regular hexagon

2b ---> the length side of the equilateral triangle

step 1

Find the area of the six triangles

Multiply the area of one triangle by 6

A=6[x^{2}\frac{\sqrt{3}}{4}]

A=3x^{2}\frac{\sqrt{3}}{2}

we have

x=2b\ units

substitute

A=3(2b)^{2}\frac{\sqrt{3}}{2}\\\\A=6b^{2}\sqrt{3}\ units^2

step 2

Find the area of the regular hexagon

Remember that, a regular hexagon can be divided into 6 equilateral triangles

so

The area of the regular hexagon is the same that the area of 6 equilateral triangles

A=3x^{2}\frac{\sqrt{3}}{2}

we have

x=b\ in

substitute

A=3(b)^{2}\frac{\sqrt{3}}{2}

step 3

To find out what fraction of the total area of the six triangles is the area of the hexagon, divide the area of the hexagon by the total area of the six triangles

3(b)^{2}\frac{\sqrt{3}}{2}:6b^{2}\sqrt{3}=\frac{3}{2} :6=\frac{3}{12}=\frac{1}{4}

3 0
3 years ago
"Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an acute triangle.
Contact [7]
A, b, c - the lengths of the sides of the triangle
and a ≤ b ≤ c
then:
a + b > c and if the triangle is an acute triangle then a² + b² > c².

1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \  x\\5^2+10^2 \ \textgreater \  x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x^2 \ \textless \  125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x \ \textless \  \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \  10\\ 5^2+x^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x^2 \ \textgreater \  75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x \ \textgreater \  \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}

3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \  10\\ x^2+5^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}
8 0
3 years ago
Read 2 more answers
Eight students are competing for 1st, 2nd, 3rd, and 4th violin chair in the school’s orchestra. How many different ways can all
Sidana [21]
We are tasked to solve for the number of ways the 8 students be competing for the 1st, 2nd, 3rd, 4th voilin chair in the schools orchestra. The number of ways if can only be filled is though the formula used in number counting, that is

8x7x6x5

Therefore, there are 1680 ways 
5 0
3 years ago
Read 2 more answers
A rectangle has a width of 2xy3 and a length of 4x5y6. What is the area of the rectangle?
astraxan [27]

Answer:

6x²5y²6 cm²

Step-by-step explanation:

(2xy3) x (4x5y6)

(2x) x (4x) = 6x²

(y) x (5y) =5y²

6x²5y²6 cm²

7 0
2 years ago
Question In Picture, Thanks.
Ainat [17]
6z to the 4th power - 2z to the 3rd power
5 0
3 years ago
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