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iragen [17]
3 years ago
11

What is the x intercept od this piecewise function? A) (0,2) B) (2,0) C) (0,4) D) (4,0)​

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer: Choice B) (2,0)

======================================

Set the first piece equal to zero and solve for x

x^2 - 4 = 0

x^2 = 4

x = sqrt(4) or x = -sqrt(4)

x = 2 or x = -2

Keep in mind that the first piece y = x^2-4 is only graphed when x = 2 or larger, so we ignore x = -2. This is one x intercept, but there may be more. Let's check the other graph to see what we get.

----------

Set the second piece equal to zero and solve for x

x-2 = 0

x-2+2 = 0+2

x = 2

We get the same result as above in the prior section. Because of this, the two pieces connect at this junction point.

The x intercept x = 2 leads to the location (2,0). The x intercept always occurs when y = 0.

The graph below shows this.

side note: the red piece y = x^2 - 4 looks linear, but it's actually not a straight line. It's just a really stretched out curve.

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If D is the center of the circle below and AC measures 48, what is the measure of ABC
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Answer:

24

Step-by-step explanation:

How do we know that a circle must complete up to 180, then we have to:

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Now knowing this distance we can also know the other, starting from the same concept, in addition to what the circle below is like, a semicircle refers to so we divide by 2:

(180-132) / 2 = 24

Therefore the answer is 24.

We can also do it ABC is half of ADC. So ABC = 48/2 = 24

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3 years ago
Y= x-8<br> y= 8 +x<br> What is the solution
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Step-by-step explanation:

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If
baherus [9]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: cos 330 = \frac{\sqrt3}{2}

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}

Proof LHS → RHS:

LHS                          cos 165

Double-Angle:        cos (2 · 165) = 2 cos² 165 - 1

                             ⇒ cos 330 = 2 cos² 165 - 1

                             ⇒ 2 cos² 165  = cos 330 + 1

Given:                        2 \cos^2 165  = \dfrac{\sqrt3}{2} + 1

                              \rightarrow 2 \cos^2 165  = \dfrac{\sqrt3}{2} + \dfrac{2}{2}

Divide by 2:               \cos^2 165  = \dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}

                             \rightarrow \cos^2 165  = \dfrac{2\sqrt3+4}{8}

Square root:             \sqrt{\cos^2 165}  = \sqrt{\dfrac{4+2\sqrt3}{8}}

Scratchwork:            \cos^2 165  = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2

                             \rightarrow \cos 165  = \pm \dfrac{\sqrt3+1}{2\sqrt2}

             Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

                             \rightarrow \cos 165  = - \dfrac{\sqrt3+1}{2\sqrt2}

LHS = RHS \checkmark

4 0
3 years ago
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