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Inga [223]
3 years ago
14

How much lower was the 6am temperature in Juneau than in Los Angeles

Mathematics
1 answer:
adell [148]3 years ago
6 0
A lot like shoot i think 20 or more
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A squirrel runs 23 feet down a hill to eat an acorn. Then it runs 23 feet up the hill. Write the integer that represents the squ
sdas [7]
The hill means it was already 23 feet above ground and he went and back up so the answer is 23 positive
4 0
3 years ago
Read 2 more answers
Can anyone answer this please i need to be for sure I’m right
Genrish500 [490]
The answer to this question is B, (-infinity, -28]. 

We can get this answer by first multiplying each side of the inequality by 7. That would get rid of the fraction. When one does that, the result is d + 28 \leq 0. That means that d \leq -28. In interval notation, which is the notation the problem is asking us, that would be (-infinity, -28], since d is all values less than -28 this includes infinity, but it also includes -28, so there is a ] around it. That means that the answer to this question is B, (-infinity, -28]. 
7 0
3 years ago
If a restaurant bills totals $85, what amount should be left as a 15$ tips
IgorC [24]
If the total bill is $85, and you want to leave a 15% tip, then you can calculate it as follows:

85 * 15%

= 85 * 15/100

= 85 * 0.15

= 12.75

The amount that should be left as a tip is D. $12.75.
6 0
3 years ago
Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0
2 years ago
If you actually know plz give an explanation to the answer so i know you aren't just answering for points, please help this is d
tatuchka [14]
Scientifically, the answer should be C
6 0
3 years ago
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