Find the area of the following figure.

Mathematics

2 answers:

Jet001 [13]3 years ago

7 0

The answer is: 60 sq. units

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

$\large \boxed{\text{60 units}^{2}}$

Step-by-step explanation:

You haven't given us a diagram of your figure. Assume it is like the one in the image below.

The formula for the area of a parallelogram is

A = bh

1. Determine the height of the parallelogram

Notice that the angle is 45°. That makes the triangle an isosceles right triangle, so we can use Pythagoras' Theorem.

h² + h² = (6√2)²

2h² = 36 × 2 = 72

h² = 36

h = 6

2. Calculate the area of the parallelogram

$\begin{array}{rcl}A & = & bh\\A & = & 10 \times 6\\& = & \textbf{60 units}^{\mathbf{2}}\\\end{array}\\\text{The area of the figure is $\large \boxed{\textbf{60 units}^{\mathbf{2}}}$}$

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There are 7 girls and 9 boys on a playground.

Anon25 [30]

The ratio of the number of girls to the total number of children is 7 : 16.

<h3>Ratio is defined as follows:</h3>

It is a comparison between two or more numbers that measures sizes in relation to each other
It describe how much of a thing is compared to another.

The number of girls in the playground = 7

The number of boys in the playground = 9

The total number in the playground is the sum of the number of girls and the number of boys.

Therefore,

total children in the playground = 7 + 9 = 16

The ratio of the number of girls to the total number of children can be express as follows:

Ratio = 7 : 16

learn more on ratio here: brainly.com/question/26169121?referrer=searchResults

8 0

3 years ago

Read 2 more answers

Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$