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coldgirl [10]
3 years ago
15

**BRAINLIEST 20 POINTS****

Mathematics
1 answer:
konstantin123 [22]3 years ago
8 0

Answer:

1/2

Step-by-step explanation:

Cos 60 = \frac{adjacent}{hypotenuse}

1/2 = \frac{XZ}{4}

Multiplying both sides by 4

=> XZ = 2

Sin Y = \frac{opposite}{hypotenuse}

Sin Y = \frac{2}{4}

Sin Y = 1/2

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Look at this set of ordered pairs:
forsale [732]

Answer: Yes

If I remember correctly, I believe these ordered pairs indicate a function, as there is exactly 1 output for every input (i.e., (output, input)). If there were 2 inputs of the same number, that'd make this relation not a function.

7 0
3 years ago
One - eighth of 7.92 x 10^10
liraira [26]
The answer is 9900000000 (thats 8 zeros) What you do is you find 10^10 which is 10000000000 (ten zeros) and multiply that by 7.92 to get 79200000000 then you just multiply it by 1/8 to get the answer.
3 0
3 years ago
What extra number must be included with the following list of numbers to increase the median by 1?
Anna35 [415]
The correct answer is C)18
7 0
3 years ago
If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
Someone plz help me
baherus [9]

Answer:

1.benchmark fractions

2. equivalent fractions

Step-by-step explanation:

Hope it helps!

5 0
3 years ago
Read 2 more answers
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