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8_murik_8 [283]
2 years ago
15

Select the correct answer.

Mathematics
1 answer:
steposvetlana [31]2 years ago
4 0

Answer:

C. 7

49 square rooted would be 7 and the 4 in (x+4) would be negative and not positive like the answer shows.

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Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)
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Answer:

Considering the given equation y = log_{3}x\\

And the ordered pairs in the format (x, y)

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For (\frac{1}{3}, a_{0} )

x=\frac{1}{3}

y=a_{0}

y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1

So the ordered pair will be (\frac{1}{3}, -1 )

For (1, a_{1} )

x=1

y=a_{1}

y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0

So the ordered pair will be (1, 0 )

For (3, a_{2} )

x=3

y=a_{2}

y = log_{3}x\\y = log_{3}3\\y = 1

So the ordered pair will be (3, 1 )

For (9, a_{3} )

x=9

y=a_{3}

y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2

So the ordered pair will be (9, 2 )

For (27, a_{4} )

x=27

y=a_{4}

y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3

So the ordered pair will be (27, 3 )

For (81, a_{5} )

x=81

y=a_{5}

y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4

So the ordered pair will be (81, 4 )

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Given f(x) = x^2+4, does there exist an x-intercept between [-2, 3]. Why or why not. Be sure to include in your discussion the I
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The x-intercepts are the values of x for which f(x) = 0.
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