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sattari [20]
4 years ago
14

Please help will give brainlyest 8th grade math please help need in 1min

Mathematics
1 answer:
Pavel [41]4 years ago
8 0

Answer:

I just did this yesterday, im in 7th grade but it’s still the same question. The answer is 8/4, but I would suggest writing 2 because the teacher might count it incorrect if you put 8/4.

Step-by-step explanation:

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7/12 14/24 21/36 28/48 35/60
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Which fraction has a terminating decimal as its decimal expansion?
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'One third' as a decimal is 1 ÷ 3 = 0.333333..... ⇒ The decimal is recurring

'One fifth' as decimal is 1 ÷ 5 = 0.2 ⇒ The decimal terminates at the tenth value

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'One ninth' as decimal 1 ÷ 9 = 0.111111.... ⇒ The decimal is recurring 

Answer: one fifth
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If 75% of a number is 15, 15% of that number
natta225 [31]

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3

Step-by-step explanation:

0.75 x ? = 15

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15% of 20 = 3

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3 years ago
Sophie has 5 pieces of string that are each 5 feet long. Cooper has 4 pieces of string that are each
Reil [10]

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Step-by-step explanation:

5 0
3 years ago
Given f(x) =
sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = \left\{        \begin{array}{ll}            2\cos(\pi x) \text{ for }  x \leq -1 \\ \\          \displaystyle   \frac{2}{\cos(\pi x)}\text{ for } x > -1        \end{array}    \right.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

8 0
3 years ago
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