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Genrish500 [490]
3 years ago
15

Suppose a triangle has sides a, b, and c, and the angle opposite the side of length b obtuse. What must be true?​

Mathematics
2 answers:
kotykmax [81]3 years ago
8 0

Answer:

C

Step-by-step explanation:

If the angle opposite b is the obtuse angle then the answer is a² + c² < b² because in an obtuse triangle, the sum of the squares of the shortest sides must be less than the square of the longest side. In this case, the longest side is the one opposite the largest angle, which is b.

zheka24 [161]3 years ago
5 0

Answer:

C.

Step-by-step explanation:

In triangles, the largest angle is always opposite from the longest side.

In other words, since the angle opposite to side b is obtuse and there are only 180 degrees in a triangle, then side b must be the longest side.

In an obtuse triangle, the longest side squared will always be greater than the squared values of the two shorter sides.

The correct answer is C.

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scoundrel [369]
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 640 people

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2 years ago
What are the solutions to x2 – 10x = 39?
oee [108]

Answer: x=−3 or x=13

Step-by-step explanation:

Step 1: Subtract 39 from both sides.

x2−10x−39=39−39

x2−10x−39=0

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(x+3)(x−13)=0

Step 3: Set factors equal to 0.

x+3=0 or x−13=0

x=−3 or x=13

7 0
2 years ago
Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
jeka94
\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

F_x=M^*
F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx
F=(7x^5y-2xy^3)e^{2x}+f(y)

F_y=N^*
(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}
f'(y)=0
\implies f(y)=C

The general solution is then

F(x,y)=(7x^5y-2xy^3)e^{2x}=C
7 0
3 years ago
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