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tankabanditka [31]
3 years ago
11

In the equation shown, what is the value of x?​

Mathematics
1 answer:
Zolol [24]3 years ago
5 0

Answer:

5^8/5^x=5^6 the asnwer is 2

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james wants to find out the favorite movie of students in his school. Which method would give James the BEST results? A)survey e
kolezko [41]

Answer:

D) Survey 100 randomly selected students from the school

Step-by-step explanation:

This gives an equal chance to everybody, so it's not biased and it is of everybody that goes to his school. A is only the people on the football team so its not students in his school. Walking to school has nothing to do with the survey topic and so thats also definitely not it. C again doesn't give a fair chance to everybody, so it would be possible to get an accurate, unbiased answer. D is definitely your answer!

I hope that helps!!

Have an awesome day! :)

8 0
3 years ago
Read 2 more answers
Amber went on a road trip over winter vacation. She recorded how far she had traveled in the table below in relation to the numb
svlad2 [7]

Answer:

B

Step-by-step explanation:

it shows every 4 hours, so:

560-336=224

224:4=56 per hour

6 0
3 years ago
If Dave drives at a steady speed of 50 mph, how long will it take him to drive 10 miles?
ira [324]
50 miles = 1 hour
10 miles = 1/50 x 10
= 1/5 hours
= 12 minutes

It will take him 12 minutes to drive 10 miles.
5 0
4 years ago
Marley had $495.66 in her bank account after she paid her monthly bills. She took her sister out to lunch, and paid $38.90. How
SOVA2 [1]

Answer: the answer is $456.76

Step-by-step explanation: All you have to do is subtract 38.90 from 495.66 and get your answer

5 0
3 years ago
How do you simplify that question? For algebra 1.
motikmotik
\left( \sqrt{x^{-3}} \right)^5 =  \left( \sqrt{ \frac{1}{x^3} } \right)^5= \frac{1}{ (\sqrt{x^3})^5 } = \frac{1}{ \sqrt{x^{15}} } = \frac{1}{x^7 \sqrt{x} }


<span>Explanation for the last step:</span>
\sqrt{x^{15}}= \sqrt{x^{14}*x}= \sqrt{x^{14}}* \sqrt{x} =x^{ \frac{14}{2} }* \sqrt{x} =x^7 \sqrt{x}

7 0
3 years ago
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