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lbvjy [14]
3 years ago
8

∠AZB and ∠CZB are supplementary. The m∠AZB = 2x - 4 and m∠CZB = 8x + 4. What is the measure of each angle?

Mathematics
2 answers:
g100num [7]3 years ago
7 0
The answer is A because when you add the angles to get they equal 180. Then just solve for x.

The drawing isn't to scale

timofeeve [1]3 years ago
3 0
The correct answer is <span>A. m∠AZB = 32° and m∠CZB = 148°</span>
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Answer:

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What are the intercepts of y=8x-18
swat32

Answer:

x-intercept=\frac{9}{4}

y-intercept=-18

Step-by-step explanation:

The y-intercept is where x is equal to 0, and the x-intercept is where y is equal to 0, so:

x-intercept=(x,0)\\y-intercept=(0,y)

The equation is written in slope-intercept form:

y=mx+b

m is the slope and b is the y-intercept, so the y-intercept is -18.

Now set the value of y to 0 to find the intercept of x:

0=8x-18

Solve for x. Add 18 to both sides:

0+18=8x-18+18\\18=8x

Divide both sides by 8:

\frac{18}{8}=\frac{8x}{8}  \\\\\frac{18}{8} =\frac{9}{4}\\ \\x=\frac{9}{4}

The x -intercept is \frac{9}{4}.

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7 0
3 years ago
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At an angle of elevation 610
Vadim26 [7]

Answer:

The height of the cliff CD is approximately 539.76 m

Step-by-step explanation:

The given parameters are;

The first angle of elevation with which the captain sees the person on the cliff = 61°

The second angle of elevation with which the captain sees the person on the cliff after moving 92 m closer to the cliff = 69°

The angle made by the adjacent supplementary angle to the second angle of elevation = 180° - 69° = 111°

∴ Whereby, the rays from the first and second angle of elevation and the distance the ship moves closer to the cliff forms an imaginary triangle, we have;

The angle in the imaginary triangle subtended by the distance the ship moves closer to the cliff = 180° - 111° - 61° = 8°

By sine rule, we have;

AB/(sin(a)) = BC/(sin(c))

Which gives;

92/(sin(8°)) = BC/(sin(61°))

BC = (sin(61°)) × 92/(sin(8°)) ≈ 578.165 m

BC ≈ 578.165 m

The height CD = BC × sin(69°)

∴  The height of the cliff CD = 578.165 m × sin(69°) ≈ 539.76 m.

The height of the cliff CD ≈ 539.76 m.

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