A pythagoras triplet is a set of three numbers ... not just any three numbers,
but a set where the (square of one of them) is the sum of the (squares of the
other two).
If they're related in that way, then they can be the lengths of the sides of a
right triangle.
If they're not, then they can't.
Answer:
The first one would be m^30n^20p^30
And the second one would be a^2 / d^4
Step-by-step explanation:
For 1, you have to multiply the inner exponents by the outer one so you get (m^12n^8) times (m^18n^12p^30). Then you mutiply the two together, so you add the exponents, where you get m^30n^20p^30.
For 2, since you're dividing, you have to subtract the exponents so you get a^2d^-4, but since you shouldn't have the negative you turn it into a^2 / d^4
Hope this helps :)
Answer:
h<-8/3
Step-by-step explanation:
collect like terms
-10h+7h>9-1
3h>8
divide both sides of the inequality by -3 and flip the inequality sign
h<-8/3
<h3>
Answer: C) 6</h3>
====================================================
Explanation:
The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.
It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.
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If n = 1, then we have
(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5
Let A = -5 as we'll use it later.
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If n = 2, then
(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8
Let B = 8 since we'll use this later as well
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If n = 3, then
(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11
Let C = -11
-------------------
If n = 4, then
(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14
Let D = 14.
--------------------
We'll add up the values of A,B,C,D to get the final answer
A+B+C+D = -5+8+(-11)+14 = 6
This means that

Answer:
84m2
Step-by-step explanation: