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Tresset [83]
2 years ago
13

Find the measurement of the missing side in each right triangle.

Mathematics
2 answers:
kirza4 [7]2 years ago
8 0

Answer:

12

Step-by-step explanation:

a^{2} + b^{2} = c^{2}

in this case its

5^{2} + b^{2} = 13^{2}

25 + b^{2} = 169 subtract 25 from both sides

b^{2} = 144 take square root of each side

b = 12

maria [59]2 years ago
7 0

Answer:

12

Step-by-step explanation:

----------------------------------------

In order to find the missing side of the triangle, we would need to use the Pythagorean theorem: a^2+b^2=c^2

So,

5^2+b^2=13^2

25+b^2=169

b^2=144

b=12

--------------------

Hope this helps.

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Solve equation <br><br> r/3-4=11<br><br> r = 45 <br><br> r = 35<br><br> r = 33<br><br> r = 28
Molodets [167]

Answer:

r=-11

Step-by-step explanation:

7 0
3 years ago
Please could you find the answers to the questions in the attachment.
Fudgin [204]
(\frac{x1+x2}{2} , \frac{y1+y2}{2})we need 3 equations
1. midpoint equation which is  (\frac{x1+x2}{2} , \frac{y1+y2}{2}) when you have 2 points

2. distance formula which is D= \sqrt{(x2-x1)^{2}+(y2-y1)^{2}}

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height


so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=(\frac{11+19}{2} , \frac{10+6}{2})
midpoint=(\frac{30}{2} , \frac{16}{2})
midpoint= (15,8)

point x=(15,8)



y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=(\frac{5+21}{2} , \frac{8+0}{2})
midpoint=(\frac{26}{2} , \frac{8}{2})
midpoint=(13,4)

Y=(13,4)



legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= \sqrt{(19-11)^{2}+(6-10)^{2}}
D= \sqrt{(8)^{2}+(-4)^{2}}
D= \sqrt{64+16}
D= \sqrt{80}
D= 4 \sqrt{5}
BC=4 \sqrt{5}





X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= \sqrt{(13-15)^{2}+(4-8)^{2}}
D= \sqrt{(-2)^{2}+(-4)^{2}}
D= \sqrt{4+16}
D= \sqrt{20}
D= 2 \sqrt{5}
XY=2 \sqrt{5}


the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC=4 \sqrt{5} and XY=2 \sqrt{5}

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= \sqrt{(21-5)^{2}+(0-8)^{2}}
D= \sqrt{(16)^{2}+(-8)^{2}}
D= \sqrt{256+64}
D= \sqrt{320}
D= 4 \sqrt{2}
AD=4 \sqrt{2}


so we have
AD=4 \sqrt{2}
BC=4 \sqrt{5} 
XY=2 \sqrt{5}

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
(4 \sqrt{2}+4 \sqrt{5}) times 1/2 times 2 \sqrt{5} =
(4 \sqrt{2}+4 \sqrt{5}) times \sqrt{5} [/tex] =
4 \sqrt{10}+4*5=4 \sqrt{10}+20=80 \sqrt{10}=252.982


























X=(15,8)
Y=(13,4)
BC=4 \sqrt{5}
XY=2 \sqrt{5}
Area=80 \sqrt{10} square unit or 252.982 square units







7 0
3 years ago
QT 1.
Delvig [45]
Median is the middle to find it you have to organize the list.

 2, 4, 4, 5, 6, 7, 8, 8, 8.

cross them out one by one from the ends and you get 6.

your answer will be B. $6

hope this helped!

:)
6 0
3 years ago
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