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pashok25 [27]
3 years ago
14

Help me with this, please

Mathematics
1 answer:
timofeeve [1]3 years ago
6 0
2 2/3 hours the 3rd one
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What is the answer to this????
miss Akunina [59]

Answer:8

Step-by-step explanation:

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/_A and /_B are complementary angles.<br> Given that m /_A = 23 degrees, find m /_B​
OverLord2011 [107]

Answer:

67

Step-by-step explanation:

When two angles are complimentary, it means that, together, they make a 90 degree angle.

/_A + /_B must equal 90°

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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







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Find the equation of the line through point (5,1) and perpendicular to 5x−y=−3. Use a forward slash (i.e. "/") for fractions (e.
IRINA_888 [86]
Y = -1/5x + 2

A line perpendicular to another line has an opposite reciprocal slope. What opposite means is that you flip the sign (from - to + or from + to -) and reciprocal means that you flip the numerator and denominator (so if line A is perpendicular to line B and line As slope is -3, line Bs slope would be 1/3).

First find the slope of 5x - y = -3
Add y to both sides
5x = y - 3
Add 3 to both sides
y = 5x + 3

So the slope of the line is 5. That means the slope of of the perpendicular line is -1/5.

Plug the slope of the perpendicular line and the point (5, 1) into point slope form to solve the equation

y - y1 = m(x - x1)
y - 1 = -1/5(x - 5)
y - 1 = -1/5x + 1
y = -1/5x + 2
3 0
2 years ago
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