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Novosadov [1.4K]
3 years ago
9

For questions 8 – 10, answer the questions about tangent-tangent angles.

Mathematics
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

m(arc ZWY) = 305°

Step-by-step explanation:

8). Formula for the angle formed outside the circle by the intersection of two tangents or two secants is,

Angle formed by two tangents = \frac{1}{2}(\text{Difference of intercepted arcs})

                                                   = \frac{1}{2}(220-140)

                                                   = \frac{1}{2}(80)

                                                   = 40°

9). Following the same rule as above,

Angle formed between two tangents = \frac{1}{2}(\text{Difference of intercepted arcs})

125 = \frac{1}{2}[m(\text{major arc})-m(\text{minor arc})]

250 = [m(\text{arc ZWY})-m(\text{arc ZY})]

250 = m(arc ZWY) - 55

m(arc ZWY) = 305°

Therefore, measure of arc ZWY = 305° will be the answer.

10). m(arc BAC) = \frac{1}{2}([m(\text{arc BDC})-m(\text{arc BC})])

                          = \frac{1}{2}(254-106)

                          = \frac{148}{2}

                          = 74°

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Answer:

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Step-by-step explanation:

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3 years ago
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Find the measure of the the sides of DEF then classify it by it sides. <br> D(8,-6) E(-1,-3) F(-2,5)
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Answer:

Part a) The measure of the sides of triangle DEF are

d_D_E=\sqrt{90}\ units

d_E_F=\sqrt{65}\ units

d_D_F=\sqrt{221}\ units

Part b) Is a scalene triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have the coordinates

D(8,-6) E(-1,-3) F(-2,5)

step 1

Find the length side DE

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substitute in the formula

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step 2

Find the length side EF

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substitute in the formula

d=\sqrt{(5+3)^{2}+(-2+1)^{2}}

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step 3

Find the length side DF

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substitute in the formula

d=\sqrt{(5+6)^{2}+(-2-8)^{2}}

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step 4

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we have

d_D_E=\sqrt{90}\ units

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Is a scalene triangle, because is a triangle in which all three sides have different lengths.

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3 years ago
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azamat

Answer:

\large \boxed{y = (x - 5)^{2} + 2 }

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y = x² - 10x + 27

y = ax² + bx + c

This is the general form of the equation for a parabola.

We must convert it to the vertex form

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A and c i’d say possibly not
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