<h3>
Answer: 1</h3>
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
Answer:
let x=distance 105 mph train traveled
(456-x)=distance 85 mph train traveled
Travel time = distance/speed (equal in both directions)
x/105=(456-x)/85
85x=105*456-105x
190x=105*456
x=105*456/190=252 miles
travel time=252/105=2.4 hrs
ans:
The two trains will meet in 2.4 hrs
Step-by-step explanation:
there brainliest pls
Ya it’s 94 i’m pretty sure
Answer:
60/120/180
Step-by-step explanation:
Either of these would work because it must be an even number divisible by 2.
It must end in a zero : divisible by 5 but even.
It must be a multiple of 30 as 3 only goes into 10s at 30. And that takes care of the 6 too.
But 4 doesn't go into 30, so 60?….
Yes, 60 is divisible between 2, 3, 4, 5 and 6
But the question does not limit your answer to one solution.
I hope this helps! :)
Answer:
A) Both estimates are slightly larger, so it is reasonable.
Step-by-step explanation:
Since 1/2 = 5/10
And 1/5 = 8/40
<u>Both estimates are larger</u>
