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Nina [5.8K]
3 years ago
7

What is the tangent ratio for angle A?

Mathematics
1 answer:
DanielleElmas [232]3 years ago
6 0
Hi there!

The trigonometry ratio tangent is equal to the side opposite the angle over the side adjacent to the angle.

The side that is opposite to angle A is side BC, which is equal to 2.
The side that is adjacent to angle A is side AC, which is equal to 3.

Therefore, the tangent ratio for angle A is 2/3. The correct answer is option A, 2/3.

Hope this helps!! :)
If there's anything else that I can help you with, please let me know! 
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Step-by-step explanation:

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144pi/8 = 18pi

part b:

total circumference = 12*2*pi = 24pi

24pi/8 = 3pi

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How to solve for axis of symmetry for y=(x+2)(x-4)
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Explain the steps to finding the vertex of g(x) = 3x2 + 12x + 15
RUDIKE [14]

Answer:

The vertex of this parabola, (-2, 3), can be found by completing the square.

Step-by-step explanation:

The goal is to express this parabola in its vertex form:

g(x) = a\, (x - h)^2 + k,

where a, h, and k are constants. Once these three constants were found, it can be concluded that the vertex of this parabola is at (h,\, k).

The vertex form can be expanded to obtain:

\begin{aligned}g(x)&= a\, (x - h)^2 + k \\ &= a\, \left(x^2 - 2\, x\, h + h^2\right) + k = a\, x^2 - 2\, a\, h\, x + \left(a\,h^2 + k\right)\end{aligned}.

Compare that expression with the given equation of this parabola. The constant term, the coefficient for x, and the coefficient for x^2 should all match accordingly. That is:

\left\lbrace\begin{aligned}& a = 3 \\ & -2\,a\, h = 12 \\& a\, h^2 + k = 15\end{aligned}\right..

The first equation implies that a is equal to 3. Hence, replace the "a\!" in the second equation with 3\! to eliminate \! a:

(-2\times 3)\, h = 12.

h = -2.

Similarly, replace the "a" and the "h" in the third equation with 3 and (-2), respectively:

3 \times (-2)^2 + k = 15.

k = 3.

Therefore, g(x) = 3\, x^2 + 12\, x + 15 would be equivalent to g(x) = 3\, (x - (-2))^2 + 3. The vertex of this parabola would thus be:

\begin{aligned}&(-2, \, 3)\\ &\phantom{(}\uparrow \phantom{,\,} \uparrow \phantom{)} \\ &\phantom{(}\; h \phantom{,\,} \;\;k\end{aligned}.

8 0
3 years ago
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