Answer: 1ST ONE CUZ THE LINE IS AT A CERTAIN ANGLE AT A POINT TO MATCH THE FIRST ANSWER
Step-by-step explanation:
sorry dude I'm stupid I would help but I'm a idot
Problem 1)
The base of the exponential is 12 which is also the base of the log as well. The only answer choice that has this is choice B.
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Problem 2)
log(x) + log(y) - 2log(z)
log(x) + log(y) - log(z^2)
log(x*y) - log(z^2)
log[(x*y)/(z^2)]
Answer is choice D
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Problem 3)
log[21/(x^2)]
log(21) - log(x^2)
log(21) - 2*log(x)
This matches with choice B
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Problem 4)
Ln(63) = Ln(z) + Ln(7)
Ln(63)-Ln(7) = Ln(z)
Ln(63/7) = Ln(z)
Ln(9) = Ln(z)
z = 9
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Problem 5)
Ln(5x-3) = 2
5x-3 = e^2
5x = e^2+3
x = (e^2+3)/5
This means choice A is the answer
Im not sure for B but A makes a square. If you have graph paper you can plot them and it should make a normal square. Hope that helps.
Answer:
A
Step-by-step explanation:
Using Pythagoras' identity on the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
t² + 12² = 13²
t² + 144 = 169 ( subtract 144 from both sides )
t² = 25 ( take the square root of both sides )
t = 
= 5 → A
