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3 years ago

Solve 6 sin θ − 3 csc θ = 0

2 answers:

8 0

$6 \sin \theta - 3 \csc \theta=0

6\sin ^2 \theta - 3=0

2 \sin ^2 \theta -1=0

2 \sin ^2 \theta =1 

\sin ^2 \theta = \frac{1}{2}

\sin \theta = \sqrt{\frac{1}{2}}

\sin \theta = \frac{{\sqrt{2}}}{2}

$

Setler79 [48]3 years ago

4 0

No solution is the answer

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Which ordered pair is the solution to the system of inequalities

Kryger [21]

Answer:

C) $(9, 4)$

Step-by-step explanation:

$4 < 5$ ☑

$4 < 3[9] → 4 < 27$ ☑

Therefore, this solution is genuine.

I am joyous to assist you anytime.

3 0

3 years ago

Find the measure of angle A when two angles are congruent?

erica [24]

m∡A = 70º

1) Considering that the Sum of the interior angles within a triangle is always 180º

2) We can write the following, and solve for x

x+80 + x +80 +40 = 180

2x +160 +40 = 180

2x + 200 = 180

2x =180-200

2x= -20

x=-10

3) So since angle A = x +80, we can plug it into that the value of x

m∡A = x +80º

m∡A=-10+80

m∡A = 70º

4 0

1 year ago

It is an odd number. it is greater than 142. it is less than 156. rounded to the nearest ten, the rounded value is less than the

GaryK [48]

The odd numbers between 142 and 156 are these:

143, 145, 147, 149, 151, 153, and 155.

Of these, 143 to the nearest ten is 140.

Likewise, 151 to the nearest ten is 150.

And, 153 to the nearest ten is 150.

Of the seven original numbers, 143, 151, and 153 round to a nearest ten that
is less than the possible mystery number.

To the nearest hundred, 143 goes to 100; 151 goes to 200, and 153 goes to 200.

Of those three, only two round to a nearest hundred that is bigger than the mystery number.

Answer: 151 and 153 are the mystery numbers.

5 0

3 years ago

What is the ratio of 12m to 10m

Hoochie [10]

Since units are same, they can be disregareded for now
12 to 10=12:10
treat as fraction
12/10=6/5
ratio is 6:5

4 0

3 years ago

Assuming boys and girls are equally likely, find the probability of a couple having a baby girl when their sixth child is born

Ivanshal [37]

Answer: The required probability of having 6th girl is 0.5.

Step-by-step explanation: Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B} and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.$

Thus, the required probability of having 6th girl is 0.5.

3 0

3 years ago