Answer:
-9
Step-by-step explanation:
The other numbers are floating point numbers
The answer is I think √10 ≈ 3.16
Answer:
The product is positive, thus it is ![\bold{+7a^4b^3}](https://tex.z-dn.net/?f=%5Cbold%7B%2B7a%5E4b%5E3%7D)
Step-by-step explanation:
The full question in proper notation is:
"Find the product. If the result is negative, enter "-". If the result is positive, enter "+".
"
We have to work with it using Order of operations know as well as PEMDAS, thus expression inside parenthesis go first and exponents.
On this expression we have to work with exponents
![(-a^2)^2 = (-a^2)(-a^2) =a^4](https://tex.z-dn.net/?f=%28-a%5E2%29%5E2%20%3D%20%28-a%5E2%29%28-a%5E2%29%20%3Da%5E4)
Thus we get
![-7(-a^2)^2(-b^3)=-7a^4(-b^3)](https://tex.z-dn.net/?f=-7%28-a%5E2%29%5E2%28-b%5E3%29%3D-7a%5E4%28-b%5E3%29)
Lastly we can work with multiplication and remembering that the multiplication of two negative signs becomes positive.
![-7(-a^2)^2(-b^3)=7a^4b^3](https://tex.z-dn.net/?f=-7%28-a%5E2%29%5E2%28-b%5E3%29%3D7a%5E4b%5E3)
So the final simplified expression is ![\bold{7a^4b^3}](https://tex.z-dn.net/?f=%5Cbold%7B7a%5E4b%5E3%7D)
Answer: 9 1/3 - 2/3 = 8 2/3 .
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Note: If the answer were "8 1/3" ;
then: "8 1/3 + 2/3 =? 9 1/3 ? "
→ "8 1/3 + 2/3 = 8 3/3 = 8 + 1 = 9 = only "9" ;
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So the answer has to be MORE than "8 1/3"
Try "8 2/3" → "8 2/3 + 2/3 =? 9 1/3?" ?? ;
→ "8 2/3 + 2/3 = 8 4/3 = 8 + 1 1/3 = 9 1/3 " → Yes!
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So, the answer is: "8 2/3" .
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Another method:
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Given the problem: "<span>9 1/3 - 2/3 " ;
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Note that: "2/3 = 1/3 + 1/3"
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So, "</span><span>9 1/3 - 2/3 = 9 1/3 - (1/3 + 1/3) = 9 1/3 - 1/3 - 1/3 = ?
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Start with: "</span>9 1/3 - 2/3" = 9. Then 9 - 1/3 = 8 3/3 - 1/3 = 8 2/3.
<span>______________________________________________________
</span>Another method:
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Given the problem:
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"9 1/3 - 2/3 = ?? " ;
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Convert "9 1/3" into "28/3" ; ("3*9 = 27"); ("27+1=28");
(The "3" comes from the "3" in the: "1/3" portion.)
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So, we rewrite as:
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28/3 - 2/3 = (28 - 2) / 3 = 26/3 ; or, 8 2/3 .
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Answer:
3.3333 hours
Step-by-step explanation:
From the attached table :
Downhill skiing for 1/6 hours ; results in 100 calories burn
To burn 1000 calories :
Skiing hours required :
(1000 / 100) * (1/6)
10 * 1/6 = 10 /6 hours
Planned skiing hours = 5
Number of skiing hours after lunch will be :
5 hours - 10/6 hours
5 hours - 1.6666 hours
= 3.3333 hours