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4 years ago

5

PLEASE HELP IM LITERALLY SO STRESSED RIGHT NOW. pleaseeeeeeee :(((

1 answer:

AlexFokin [52]4 years ago

6 0

B is the right answer because you can set up a ratio like this:
$tan(A) = (\frac{16.3}{16.3})$

If you solve for A, then you get:
$A = tan^-1(\frac{16.3}{16.3})$

Hopefully this makes sense. Also, in the textbox, ∠A = 45° and your answer's B.

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Gnesinka [82]

Answer:

a

because 1 1/2 * 1 1/2 *1 = 2 1/4

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3 years ago

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Which one of these is the correct unit of measurement in finding volume?

kirill [66]

Answer:

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Step-by-step explanation:

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3 years ago

Solve :3(2x+1)^-1 + 5x=-1<br>

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hope this is ans . Brother

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3 years ago

Solve for x:<br> -3(x-2)^2+17=0 <br> round your answer to the nearest hundredth

kicyunya [14]

Answer:

-3(x-2)×2+17=0

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8 0

3 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%

Llana [10]

a) The given function is

$f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}$

The domain refers to all values of x for which the function is defined.

The function is defined for

$x^4+x^3-4x^2-4\ne0$

This implies that;

$x\ne -2.69,x\ne 1.83$

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

$x^4+x^3-4x^2-4=0$

This implies that;

$x= -2.69,x=1.83$

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

$\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

$\Rightarrow x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm2$

The roots are $x=-2,x=2$

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

$x=0$ into the function

$f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}$

$f(0)=\frac{-4}{-4}=1$

e) to find the horizontal asypmtote, we take limit to infinity

$lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

The horizontal asymtote is $y=0$

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

6 0

3 years ago