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antiseptic1488 [7]

2 years ago

12

What is a good comparison of the estimated sum and the actual sum of 7 9/12+2 11/12

1 answer:

earnstyle [38]2 years ago

4 0

10 8/12 because you get an improper fraction when you add 9/12 and 11/12 (20/12) so you see how many times 12 goes into 20 which is only once. you see how many are left over from when you take 12 from 20. then you have you fraction which is 8/12. now you have to add the numbers which is 7+2=9 but 12 goes into 20 once so you add 1 to your number to make it 10. That's the answer, 10 8/12.

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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)

vagabundo [1.1K]

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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Savatey [412]

We want 4/5 of a number to be 144. This means that

$\dfrac{4}{5} \cdot x = 144$

Multiplying both sides by 5/4, we have

$x = \dfrac{144\cdot 5}{4} = 180$

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