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svlad2 [7]
3 years ago
10

If ef bisects angle ceb,angle cef=7x+31 and angle feb=10x-3

Mathematics
1 answer:
Nastasia [14]3 years ago
7 0

Given : Angle  < CEB is bisected by EF.

< CEF = 7x +31.

< FEB = 10x-3.

We need to find the values of x and measure of < FEB, < CEF  and < CEB.

Solution: Angle  < CEB is bisected into two angles < FEB and < CEF.

Therefore,   < FEB = < CEF.

Substituting the values of < FEB and < CEF, we get

10x -3 = 7x +31

Adding 3 on both sides, we get

10x -3+3 = 7x +31+3.

10x = 7x + 34

Subtracting 7x from both sides, we get

10x-7x = 7x-7x +34.

3x = 34.

Dividing both sides by 3, we get

x= 11.33.

Plugging value of x=11.33 in < CEF = 7x +31.

We get

< CEF = 7(11.33) +31 =  79.33+31 = 110.33.

< FEB  = < CEF =  110.33 approximately

< CEB = < FEB +  < CEF  = 110.33 +110.33 = 220.66 approximately



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What is 9/10 + 7/15
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\huge\text{Hey there!}

\mathsf{\dfrac{9}{10}+\dfrac{7}{15}}

\large\textsf{FIRST: FIND the LCD (Lowest Common Denominator) then solve}\\\large\textsf{from there!}

\large\textsf{If you have calculated it correctly, you should have came up with \underline{\bf 30}}\\\large\textsf{as your LCD (Lowest Common Denominator).}

\mathsf{= \dfrac{9\times3}{10\times3}+ \dfrac{7\times2}{15\times2}}

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\mathsf{= \dfrac{41}{30}}\large\textsf{ which you could convert to }\mathsf{1 \dfrac{11}{30}}

\boxed{\boxed{\large\textsf{ANSWER: }\bf \dfrac{41}{30} \large\textsf{ or }\mathsf{\bf 1 \dfrac{11}{30}\large\textsf{ because they both equal the same thing}}}}}\huge\checkmark

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