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Scorpion4ik [409]
3 years ago
9

Content - Common

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

9514 1404 393

Answer:

  • ∠BCA = 30°
  • ∠FHC = 125°
  • ∠A = 45°
  • ∠GBC = 75°
  • arc EBC = 260°

Step-by-step explanation:

Angle BCA intercepts arc FB, so is half its measure: 60°/2 = 30°.

Angle FHC is half the sum of the angles intercepted by its chords: (60°+150°+40°)/2 = 125°.

Angle A is half the difference of the arcs it intersects: (150° -60°)/2 = 45°.

Angle GBC is half the measure of the arc it intercepts: 150°/2 = 75°.

arc EBC is the sum of arcs EF, FB, and BC: 50° +60° +150° = 260°.

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Alex
Since 2.5% as a decimal is .025, the equation to get this is:

(.025)(80000)

Once multiplied, these equal 2,000. Or in other words, the Smiths pay $2,000 in school taxes per year.
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3 years ago
If f(x)=2x+5 and g(x)=x²+5, evaluate the functions:
Ostrovityanka [42]

Answer:

f(-3) = -1

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Step-by-step explanation:

plug -3 for x to get the answer

3 0
3 years ago
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A parabola has a vertex at (0,0). The focus of the parabola is located at (4,0). What is the equation of the directrix?
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The equation is y^2=16x and the directrix is x=-4
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3 years ago
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What is the period for the function y = -4 cos (5x +3)? (4.3)
BlackZzzverrR [31]

The periodicity of function is \frac{2 \pi}{5}

<em><u>Solution:</u></em>

Given that we have to find the period of function

<em><u>Given function is:</u></em>

y = -4 + cos(5x+3)

<em><u>Use the below formula:</u></em>

\mathrm{Periodicity\:of\:}a\cdot \cos \left(bx\:+\:c\right)\:+\:d=\frac{\mathrm{periodicity\:of}\:\cos \left(x\right)}{|b|}

Thus,

Peridocity\ of\ cos(5x + 3) - 4 = \frac{\text{periodicity of cos(x)}}{5}

Now find the periodicity of cos(x)

We know that,

periodicity of cos(x) = 2 \pi

Therefore,

Peridocity\ of\ cos(5x + 3) - 4 = \frac{2 \pi}{5}

Thus the periodicity of function is \frac{2 \pi}{5}

4 0
3 years ago
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Answer:

C x 34 or 34c

Step-by-step explanation:

There isn't really a way to show work or really an answer unless you have a value for C but I hope this helped anyway.

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