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ratelena [41]
3 years ago
9

Given: AD diameter of circle P. m AB + m BC = measure of A)minor arc AC B) major arc AC

Mathematics
2 answers:
Effectus [21]3 years ago
5 0

Answer: minor arc AC

Step-by-step explanation:

Makovka662 [10]3 years ago
3 0

In the given circle, m AB is represented by angle 1 and m BC is represented by angle 2 .

And

m \ AB + m \ BC = \angle 1 + \angle 2 = m \AC

And it represents the minor arc AC .

So the correct option is A.

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Agata [3.3K]
Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify! 

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, 3^{-2} =
\frac{1}{3^{2} }.
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, (\frac{3}{4}) ^{3}  =  \frac{ 3^{3} }{4^{3} }.
3) The zero exponent rule<span> says that any number raised to zero is 1. For example, 3^{0} = 1.
</span>

Back to the Problem:
Problem 1 
The x-values are in the left column. The title of the right column tells you that the function is y =  4^{-x}. The x-values are:
<span>1) x = 0
</span>Plug this into y = 4^{-x} to find letter a:
y = 4^{-x}\\&#10;y = 4^{-0}\\&#10;y = 4^{0}\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = 4^{-x} to find letter b:
y = 4^{-x}\\ &#10;y = 4^{-2}\\ &#10;y =  \frac{1}{4^{2}} \\  &#10;y= \frac{1}{16}
<span>
3) x = 4
</span>Plug this into y = 4^{-x} to find letter c:
y = 4^{-x}\\ &#10;y = 4^{-4}\\ &#10;y =  \frac{1}{4^{4}} \\  &#10;y= \frac{1}{256}
<span>

Problem 2
</span>The x-values are in the left column. The title of the right column tells you that the function is y =  (\frac{2}{3})^x. The x-values are:
<span>1) x = 0
</span>Plug this into y = (\frac{2}{3})^x to find letter d:
y = (\frac{2}{3})^x\\&#10;y = (\frac{2}{3})^0\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = (\frac{2}{3})^x to find letter e:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\&#10;y =  \frac{4}{9}
<span>
3) x = 4
</span>Plug this into y = (\frac{2}{3})^x to find letter f:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}
<span>
-------

Answers: 
a = 1
b = </span>\frac{1}{16}<span>
c = </span>\frac{1}{256}
d = 1
e = \frac{4}{9}
f = \frac{16}{81}
5 0
3 years ago
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