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Svetach [21]
3 years ago
8

Please could you find the answers to the questions in the attachment.

Mathematics
1 answer:
Fudgin [204]3 years ago
7 0
(\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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The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep
Alchen [17]
Ooh, fun

geometric sequences can be represented as
a_n=a(r)^{n-1}
so the first 3 terms are
a_1=a
a_2=a(r)
a_2=a(r)^2

the sum is -7/10
\frac{-7}{10}=a+ar+ar^2
and their product is -1/125
\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3

from the 2nd equation we can take the cube root of both sides to get
\frac{-1}{5}=ar
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r
subsituting -1/5 for ar
\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r
which simplifies to
\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}
multiply both sides by 10r
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add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for ax^2+bx+c=0
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
so
for 2r²-5r+2=0
a=2
b=-5
c=2

r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}
r=\frac{5 \pm \sqrt{25-16}}{4}
r=\frac{5 \pm \sqrt{9}}{4}
r=\frac{5 \pm 3}{4}
so
r=\frac{5+3}{4}=\frac{8}{4}=2 or r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}

use them to solve for the value of a
\frac{-1}{5}=ar
\frac{-1}{5r}=a
try for r=2 and 1/2
a=\frac{-1}{10} or a=\frac{-2}{5}


test each
for a=-1/10 and r=2
a+ar+ar²=\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}
it works

for a=-2/5 and r=1/2
a+ar+ar²=\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}
it works


both have the same terms but one is simplified

the 3 numbers are \frac{-2}{5}, \frac{-1}{5}, and \frac{-1}{10}
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