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serg [7]
3 years ago
6

Find the area of the figure. Round to the nearest tenth if necessary. Use 3.14 for .

Mathematics
1 answer:
Anna [14]3 years ago
5 0

Answer:

22\ units^2

Step-by-step explanation:

we know that

The area of the figure is equal to the area of a triangle plus the area of a parallelogram

<em>Find the area of triangle KLM</em>

A=\frac{1}{2} (b)(h)

we have

b=7-3=4\ units --> difference of the x-coordinates points M and K

h=6-3=3\ units --> difference of the y-coordinates points L and K

substitute

A_1=\frac{1}{2} (4)(3)=6\ units^2

<em>Find the area of parallelogram JKMN</em>

A=(B)(H)

B=6-2=4\ units --> difference of the x-coordinates points N and J

H=3-(-1)=4\ units --> difference of the y-coordinates points K and J

substitute

A_2=(4)(4)=16\ units^2

The area of the figure is equal to

A=A_1+A_2

A=6+16=22\ units^2

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Step-by-step explanation:

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2.  4v-10=-26

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Dylan drinks 2/3 of a glass of water for breakfast, 1 1/4 glasses for lunch, and 2 glasses for dinner. How much water did he dri
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Step-by-step explanation:

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A 2x2 square is centered on the origin. It is dilated by a factor of 3. What are the coordinated of the vertices of the square?
Alexxx [7]

<u>Answer</u>:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

The ratio of area of  larger square to smaller square is 9:1

<u>Step-by-step explanation:</u>

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1),  B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

To find the area of the small square

the side  of the small square is 2 units

so the are of the small square is 2^2 = 4 square units

To find the area of the larger square

lets find the side AB of the square using distance formula

=>\sqrt{(x_2 -x_1)^2 +(y_2-y_1)^2}

=>\sqrt{(3 - (-3))^2 +(-3 - (-3))^2}

=>\sqrt{(3 +3)^2 +(-3 +3)^2}

=>\sqrt{(6)^2 +(0)^2}

=>\sqrt{36}

=>6

AB =6 units

In a square all the sides will be equal

Now the area of the larger square will be

6^2

36 square units

The ratio of larger square to smaller square is

=>36 : 4

=>9 : 1

5 0
3 years ago
For five mathematics tests, your scores were 81, 86, 81, 76, 71. what was your average score?
trasher [3.6K]

add them together then divide by 5

81 +86 +81 +76 +71 = 395

395/5 = 79

 your average was 79

7 0
3 years ago
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