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Scilla [17]
3 years ago
5

Hurry pls need help look at picture below

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0
V = S*h
h = V/S = 250*pi / 50*pi = 5 m 

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3 years ago
Pretest: Unit 1
Alex_Xolod [135]

A vertical line that the graph of a function approaches but never intersects. The correct option is B.

<h3>When do we get vertical asymptote for a function?</h3>

Suppose that we have the function f(x) such that it is continuous for all input values < a or > a and have got the values of f(x) going  to infinity or -ve infinity (from either side of   x = a) as x goes near a, and is not defined at   x = a, then at that point, there can be constructed a vertical line  x = a and it will be called as vertical asymptote for f(x) at   x = a

A vertical asymptote can be described as a vertical line that the graph of a function approaches but never intersects.

Hence, the correct option is B.

Learn more about Vertical Asymptotes:

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3 0
1 year ago
12. What value of c would make x = 4 in the<br> equation 5 + cx = 13 ?
Pavel [41]
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4 0
3 years ago
PLEASE HELP ASAP ILL GIVE BRAINLIEST!!!!!<br><br>find measure of arc MK​
Gwar [14]

Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

\sf \cos(\theta)=\dfrac{A}{H}

where:

  • \theta is the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

Given:

  • \theta = ∠KLM
  • A = LN = 8
  • H = KL = 15.2

\implies \sf \cos(KLM)=\dfrac{8}{15.2}

\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)

\implies \sf \angle KLM=58.24313614^{\circ}

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}

Given:

  • r = 15.2
  • ∠KLM = 58.24313614°

\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)

\implies \textsf{Arc length MK}=\sf 15.45132428\:units

6 0
2 years ago
How would I find the size of angle x
ivolga24 [154]

Answer: the answer would be 7

Step-by-step explanation:

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8 0
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