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Ostrovityanka [42]
3 years ago
11

Given sin θ = square root of 23 divided by 12 and tan θ = square root of 23 divided by 11, find cos θ. Enter the answer as a fra

ction in lowest terms.

Mathematics
1 answer:
ohaa [14]3 years ago
3 0

Answer:

11/12

Step-by-step explanation:

tan θ = sin θ / cos θ

cos θ = sin θ / tan θ

cos θ = (√23/12) / (√23/11)

cos θ = (√23/12) × (11/√23)

cos θ = 11/12

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BARSIC [14]

Answer:

C. Both statements are true

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Belinda walks 10 minutes on the first day and plans to increase the number of minutes by 2 minutes each day.How many will Belind
Korolek [52]

Answer:

24 minutes

Step-by-step explanation:

This can be explained with the equation y=2x+10 (I chose 2 arbritrary variables, any are acceptable). A week is equivalent to 7 days so x=10. Plug this into the equation and you end up with 24, because 2(7)+10=24

8 0
3 years ago
Please help me!! Show work please :)
lara31 [8.8K]

Hello!

As you can see, all of these numbers just go down to the tenths, not the hundreds, so we cannot round.

Therefore, the numbers will remain the same.

I hope this helps!

8 0
3 years ago
Read 2 more answers
Write 3,047.092 in expanded form
Grace [21]

Step-by-step explanation:

\:  \:  \:  \:  \:  \:  \:  \:  \: =  >  \: 3000 \:  +  \: 40 \:  +  \: 7 \:  +  \:  \frac{9}{100}  \:  +  \:  \frac{2}{1000}

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7 0
3 years ago
A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer
Anna11 [10]

Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

L_A \to length of play area A

W_A \to width of play area A

L_B \to length of play area B

W_B \to width of play area B

x \to Area of A

y \to Area of B

From the question, we have the following:

L_A = 1 + 4W_A

W_B = 2 + W_A

L_B = 2 + 3W_B

x = y

The area of A is:

x = L_A * W_A

This gives:

x = (1 + 4W_A) * W_A

Open bracket

x = W_A + 4W_A^2

The area of B is:

y = L_B * W_B

y = (2 + 3W_B) * ( 2 + W_A)

Substitute: W_B = 2 + W_A

y = (2 + 3(2 + W_A)) * ( 2 + W_A)

Open brackets

y = (2 + 6 + 3W_A) * ( 2 + W_A)

y = (8 + 3W_A) * ( 2 + W_A)

Expand

y = 16 + 8W_A + 6W_A + 3W_A^2

y = 16 + 14W_A + 3W_A^2

We have that:

x = y

This gives:

W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2

Collect like terms

4W_A^2 - 3W_A^2 + W_A  -14W_A  - 16 =0

W_A^2  -13W_A  - 16 =0

Using quadratic calculator, we have:

W_A = -14.1 or W = 1.13 --- approximated

But the width can not be negative; So:

W = 1.19

7 0
3 years ago
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