3 years ago

What are the angles of a triangle with the sides 10, 8, 7

2 answers:

4 0

Angle ∠ A = α = 44.049° = 44°2'55″ = 0.769 rad
Angle ∠ B = β = 52.617° = 52°37' = 0.918 rad
Angle ∠ C = γ = 83.335° = 83°20'4″ = 1.454 rad

3 0

10> trapezoid
7>homagrama
8>square

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Karolina [17]

The answer is n=-4 hope this helps

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Zarrin [17]

According to my calculation it should be C: x= 1 and y = - 4
NOTE: I had it right on my online class.

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GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$ = $\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}$

= $\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}$

= $\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}$

= $\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}$

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3 years ago

Sergio039 [100]

41/120= 0.3416
34% out of 100%
.34 multiplied by 360 degrees
122.4 degrees

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3 years ago

Svet_ta [14]

My answer would be B

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3 years ago