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kolezko [41]
3 years ago
11

In this diagram the big square is divided into 7 congruent small squares and two congruent triangles. The area of the shaded squ

are is 4sq cm. What is the area of the shaded triangle?
PLS HELP. PLS TELL ME HOW YOU GOT THE ANSWER TOO!

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
5 0
Since each square will have an area of 4 cm^2 because they are congruent, then let’s find the side length of them.

Square root of 4 is two

Each side is 2 cm. Since the unshaded triangle has 3 on the base and height, then it is same to assume that the triangle has a base of 6 cm (2 * 3) and a height of 6 cm.

The formula of the area of a triangle is:
1/2 ( base times height)

Let’s plug in some values

1/2(6 * 6)

Multiply

1/2 (36)

Divide

18

Since the triangles are also congruent, then the shaded side has the same area.

The shaded triangle has an area of 8 cm ^2
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And in our example, x = minimum score needed, therefore

\begin{gathered} 0.84=\frac{x-480}{105} \\  \\ x=0.84\cdot105+480 \\  \\ x=568.2 \end{gathered}

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7 0
2 years ago
The length of a rectangle is half the width. The area is 25 square meters
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Mamont248 [21]

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(x +y)^5<br> Complete the polynomial operation
Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

\left(x+y\right)^5

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=x,\:\:b=y

=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i

so expanding summation

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

solving

\frac{5!}{0!\left(5-0\right)!}x^5y^0

=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5

=1\cdot \:1\cdot \:x^5

=x^5

also solving

=\frac{5!}{1!\left(5-1\right)!}x^4y

=\frac{5}{1!}x^4y

=\frac{5}{1!}x^4y

=\frac{5x^4y}{1}

=\frac{5x^4y}{1}

=5x^4y

similarly, the result of the remaining terms can be solved such as

\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2

\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3

\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4

\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5

so substituting all the solved results in the expression

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Therefore,

\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

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3 years ago
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