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wlad13 [49]
3 years ago
13

Please help it almost due to 15 points

Mathematics
1 answer:
Otrada [13]3 years ago
3 0

Answer:

a=30

b=15

c=3

d=30

e=10

Step-by-step explanation:

Solve for A first and continue solving for one variable at a time

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Question mark 1: Match the triangle to correct function and answer
postnew [5]

Answer:

Triangle 1 - m\angle A = 58^{\circ}, hypotenuse = 16

x \approx 8.479

Triangle 2 - m \angle A = 44^{\circ}, adjacent leg = 18

x \approx 17.382

Triangle 3 - m\angle A = 49^{\circ}, opposite leg = 12

x \approx 15.900

Step-by-step explanation:

Each triangle is solved by means of trigonometric relations:

Triangle 1 - m\angle A = 58^{\circ}, hypotenuse = 16

\cos 58^{\circ} = \frac{x}{16}

x = 16\cdot \cos 58^{\circ}

x \approx 8.479

Triangle 2 - m \angle A = 44^{\circ}, adjacent leg = 18

\tan 44^{\circ} = \frac{x}{18}

x = 18\cdot \tan 44^{\circ}

x \approx 17.382

Triangle 3 - m\angle A = 49^{\circ}, opposite leg = 12

\sin 49^{\circ} = \frac{12}{x}

x = \frac{12}{\sin 49^{\circ}}

x \approx 15.900

8 0
3 years ago
Help pls !! Im awful with math and I already failed and this is a makeup test >-<​
ahrayia [7]

Answer:

29.4 degrees

Step-by-step explanation:

a = 53 cm = opposite side to the angle

b = 94 cm = adjacent side to the angle \alpha

tan = opp/adjacent = 53/94 = 0.5638

\alpha = inverse tangent of .5638 = tan⁻¹ (.5638) = 29.4₋⁰

5 0
3 years ago
An unfair coin has probability 0.3 of landing heads. The coin is tossed seven times. What is the probability that it lands heads
solmaris [256]

Answer:

0.9176 = 91.76% probability that it lands heads at least once

Step-by-step explanation:

For each time that the coin is tossed, there are only two possible outcomes. Either it lands heads, or it does not. The probability of a toss landing heads is independent of other tosses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

An unfair coin has probability 0.3 of landing heads.

This means that p = 0.3

The coin is tossed seven times.

This means that n = 7

What is the probability that it lands heads at least once?

Either it does not lands heads, or it lands at least once. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{7,0}.(0.3)^{0}.(0.7)^{7} = 0.0824

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0824 = 0.9176

0.9176 = 91.76% probability that it lands heads at least once

5 0
4 years ago
Find the solutions to the equation below . Check all that apply . X^2-4=0
KatRina [158]

Answer:

there are 2 solutions in total

Step-by-step explanation:

x^2-4=0

x^2=4

x= ±2<em>i</em>

5 0
3 years ago
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Find the difference quotient for the function f(x)=−5x−3f(x)=−5x−3. simplify your answer as much as possible.
tino4ka555 [31]
Difference quotient = [f(x+h) - f(x)]/h

[-5(x+h) -3 - (-5x -3)]/h
[-5x -5h -3 + 5x +3]/h
[-5h]/h
-5
6 0
3 years ago
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