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jenyasd209 [6]
3 years ago
7

Is -$10 a solution? Substitute -10 into the inequality in the example to check.

Mathematics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

Yes

Step-by-step explanation:

-10 is less than 20

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Step-by-step explanation:

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2 years ago
Solving Exponential and Logarithmic Equation In exercise,solve for x or t.See example 5 and 6.
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Answer:

x=3 , x=-3

Step-by-step explanation:

e^{lnx^2} -9= 0

Add 9 on both sides

e^{lnx^2}=9

To remove base 'e' we take ln on both sides

ln e^{lnx^2}=ln9

lnx^2=ln(9)

Both sides have ln , so we make the terms equal

x^2=9

Take square root on both sides

x=+-3

x=3 , x=-3

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3 years ago
Find the area of the trapezoid
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Step-by-step explanation:

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6 0
3 years ago
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Is the triangle obtuse, acute, equilateral or right?
Stells [14]

9514 1404 393

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 ...

  \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}

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 ...

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

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.

4 0
3 years ago
Read 2 more answers
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