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Anna71 [15]
1 year ago
12

Question 3 of 10

Mathematics
1 answer:
GaryK [48]1 year ago
5 0

The correct value that equates to this expression is 12‐³². <u>Letter A</u>

.

To solve this expression, just: eliminate the parentheses and multiply the exponents among themselves;\boxed{  \large \sf (a {}^{n} ) {}^{m}  \rightarrow a {}^{n \times m}  } \\  \\

<h3>Resolution </h3>

{   = \large \sf (12{}^{-4} ) {}^{8}  }

{   = \large \sf 12{}^{-4 \times 8}   }

\pink{ \boxed{   = \large \sf 12{}^{-32}   } } \\

Therefore, the answer will be <u>12</u><u>‐³</u><u>²</u>

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Please help!!! My math teacher doesn’t explain things well.
abruzzese [7]

When you look at the graph, the X is the x value and the one on the f(x) is the y value, imagian it is on a graph and x is the steps you have to walk then imagain a ladder you have to go as high as the f(x) value says, then mark them. if they follow a pattern then write it down, if it does not then write incosentant.

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A

Step-by-step explanation:

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6 0
2 years ago
How many 3/4 foot lengths of pipe can be cut from a 7 and 1/2 foot pipe?
stiks02 [169]

Answer:

The number of pipes to be cut from total length of pipe is 10 pipes

Step-by-step explanation:

Given as :

Total length of pipe = l =  7\frac{1}{2} foot

i.e ' = \frac{15}{2}  foot

The length of pipe to be cut = l' = \dfrac{3}{4} foot

Let The number of pipes to be cut = n pipes

<u>According to question</u>

The number of pipes to be cut = \dfrac{\textrm Total length of pipe}{Length of pipe to be cut}

i.e n = \dfrac{l}{l'}

Or, n = \frac{\frac{15}{2}}{\frac{3}{4}}

Or, n = \frac{15\times 4}{3\times 2}

or, n = 5 × 2

∴ n = 10 pipes

So, The number of pipes to be cut = n = 10 pipes

Hence, The number of pipes to be cut from total length of pipe is 10 pipes Answer

3 0
3 years ago
The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this mi
zzz [600]

Answer:

False this value is very likely to find in this distribution

Step-by-step explanation:

With mean 7 ( μ ) and standad deviation  (σ ) 2 we can observe, value 6.6  is close to the lower limit of the interval

μ  ± 0,5 σ      7 ± 1 in which we should find 68,3 % of all values

(just 6 tenth to the left)

And of course 6.6 is inside the interval

μ  ± 1 σ   where we find 95.7 % of th value

We conclude this value is not unlikely at all

4 0
3 years ago
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