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3 years ago

5

Find the area of the figure plz

2 answers:

Lana71 [14]3 years ago

4 0

The answer to the question above is exactly 72.5ft

uranmaximum [27]3 years ago

3 0

Answer:

72.5ft

Step-by-step explanation:

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23 + 12 ×13 ÷2 Math homework

Contact [7]

Answer:

101 is your answer

Step-by-step explanation:

Remember to follow PEMDAS & the left->right rule.

First, multiply 12 with 13:

12 x 13 = 156

Next, divide 156 with 2

156/2 = 78

Finally, add 23

78 + 23 = 101

101 is your answer

~

5 0

3 years ago

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Jatonia plans to purchase 4

steposvetlana [31]

Hey there!

All we need to do is multiply the unit price of the pencil ($.35) by how many pencils she wants to buy (4)

$.35 x 4 = $1.40

Therefore, she will pay $1.40 for 4 pencils

Hope this helps you!

God bless ❤️

xXxGolferGirlxXx

6 0

3 years ago

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7 lbs of sour candies cost $7.00. What is the unit price?

grandymaker [24]

Answer:

$1 per lb

Step-by-step explanation:

If you divide the cost by the amount you are getting, then you get the unit rate. So, since 7 divided by 7.00 is 1.00, then the cost per pound is $1.00

7 0

3 years ago

Divide 254 by 11.8. Round your answer to the nearest hundredth. A. 20.97 B. 21.53 C. 21.35 D. 22.07

Rainbow [258]

The answer is B 21.53

7 0

3 years ago

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

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