Is the triangle obtuse, acute, equilateral or right?

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Answer:

  obtuse

Step-by-step explanation:

The law of cosines tells you ...

  b² = a² +c² -2ac·cos(B)

Substituting for a²+c² using the given equation, we have ...

  b² = b²·cos(B)² -2ac·cos(B)

We can subtract b² to get a quadratic in standard form for cos(B).

  b²·cos(B)² -2ac·cos(B) -b² = 0

Solving this using the quadratic formula gives ...

  [tex]\cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}[/tex]

The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...

  [tex]\cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}[/tex]

The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.

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Give a good explanation

lets evaluate 81^-3/4.

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Initial value: 1, rate of change 3 over 5.
Initial value: 3 over 5., rate of change: 0
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does this make sence to anyone

simplify

(x^a.x^b)^a-b*(x^b.x^c)^b-c*(x^c.x^a)^c-a

step by step..

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Help me please! It’s due in like 10 minutes and I’m barely on the 8th question !!!

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15 POINTS!!!
Identify the initial value and rate of change for the graph shown

A coordinate plane graph is shown. A line passes through the y-intercept at 1 and through the point 5 comma 4.

Initial value: 1, rate of change 3 over 5.
Initial value: 3 over 5., rate of change: 0
Initial value: 1, rate of change 5 over 3.
Initial value: 5 over 3., rate of change: 0

HELP PLEASE!!! i will give brainly! +10 points

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