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2 years ago

Use the circle to solve for the missing measures. Assume lines that appear tangent are tangent. Do NOT round any answers.

Mathematics

1 answer:

wlad13 [49]2 years ago

4 0

The given circle having two external tangents, according to circle

theorem, have the following values;

r = 3

m∠1 = 82°

m∠2 = 49°

x = 82°

<h3>How can the radial length and angles be found?</h3>

r² + 1.6² = (0.4 + r)²

(0.4 + r)² - (r² + 1.6²) = 0

0.8·r - 2.4 = 0

0.8·r = 2.4

r = 2.4 ÷ 0.8 = 3

r = 3

y = 49°

From the two tangent angle theorem, we have;

m∠1 = ((82 + 98 + 82) - (360 - ((82 + 98 + 82)))) ÷ 2 = 82

m∠1 = 82°

According to circle theorems, angle at the center is twice angle

subtended at the circumference.

m∠2 = (360 - (82 + 98 + 82)) ÷ 2 = 49°

m∠2 = 49°

The length of the chord subtended by the arc x is equal to the length of

the chord subtended by 89°, which gives;

x = 82°

Learn more about circle theorems applications here:

brainly.com/question/16879446

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Ldentify the following sections of the textbook.

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3 years ago

Line AB passes through points A (-6,6) and B(12,3). If the equation of the line is written in slope-intercept form, y=mx+b, the.

Tomtit [17]

Answer:

You wrote the question wrong...

m is not -16 it is -1/6 (that is a big difference)

y=mx+b

y=-1/6x+b

3 = -1/6(12) + b

3 = -2 + b

b=5

Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

3 years ago

ASAP HELP RN

Akimi4 [234]

Answer:

Part B: is -60 divided by -10

Part C: is The number of increments is positive.

Part D: is -1 times 60 / -1 times 10

Part E: is Any number divided by itself is 1. In the expression -1 times 60, the -1s in the numerator and the denominator cancel each other out to give the result 60/10, or 6

Part F: is There are 6 increments of -10 feet in -60 feet.

Part G is: copy and paste -10 feet 6 times

Part H: is -60 / 4 increments

Part I: is there is one negative number in the expression, so the solution will be negative. There isn't a -1 in the denominator to cancel out the -1 in the numerator.

Part J: is -60 /4 =-15 Alex’s elevation change in each increment was -15 feet.

Step-by-Step Explaination: I did the test.

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3 years ago