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mel-nik [20]
3 years ago
7

Study the diagram of circle B. Points C, R, and V lie on circle B. The radius, BC¯¯¯¯¯¯¯¯, and the diameter, VR¯¯¯¯¯¯¯¯, are dra

wn. Arc CR has a measure of 90∘. If m∠VBC=(3x+4)∘, what is the value of x?

Mathematics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

<h2>D. x ≈ 31.3°</h2>

Step-by-step explanation:

The triangle CBV is angled triangle with right angle at ∠VBC.

Since  ∠VBC = (3x-4) °

To get the value of x, we will equate the angle  ∠VBC to 90°

This results into (3x-4) ° = 90

Simplifying the resulting equation;

3x-4 = 90

Adding 4 to both sides;

3x-4+4 = 90+4

3x = 94

Dividing both sides by 3

3x/3 = 94/3

x = 31.33°

x ≈ 31.3°

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4 0
3 years ago
Verify the identity 6cos^2(x) -3 = 3 - 6sin^2(x)
Lorico [155]

Answer:

It is proved that  6\cos ^{2}x -3 = 3 - 6\sin ^{2} x .

Step-by-step explanation:

We already have the identity of x as \sin ^{2}x + \cos ^{2}x = 1 .......... (1)  .

So, from equation (1) we can write that

\cos ^{2} x = 1 - \sin ^{2} x

⇒ 6\cos ^{2} x = 6 - 6 \sin ^{2} x

⇒ 6\cos ^{2} x -3 = 6 - 6 \sin ^{2}x -3

⇒ 6\cos ^{2}x -3 = 3 - 6\sin ^{2} x

Hence, it is proved that  6\cos ^{2}x -3 = 3 - 6\sin ^{2} x . (Answer)

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3 years ago
The triangle shown below has an area of 20 units.
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3 0
3 years ago
22 - 10 + (18 ÷ 3) =
juin [17]
Follow PEMDAS (parenthesis- exponents (& roots) - multiplication - division - addition - subtraction)


18/3 = 6

22 - 10 = 12

12 + 6 = 18

18 is your answer

hope this helps
5 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
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