Answer:
x = 71
Step-by-step explanation:
x + 25 = 96
=> x = 96 - 25
=> x = 71
Answer:
Step-by-step explanation:
Remove perfect squares from under the radicals.
_____
The applicable rules of exponents are ...
(x^a)(x^b) = x^(a+b)
√(a^2) = a . . . . . . . for a > 0
(√a)(√b) = √(ab)
Given that <span>Line m is parallel to line n.
We prove that 1 is supplementary to 3 as follows:
![\begin{tabular} {|c|c|} Statement&Reason\\[1ex] Line m is parallel to line n&Given\\ \angle1\cong\angle2&Corresponding angles\\ m\angle1=m\angle2&Deifinition of Congruent angles\\ \angle2\ and\ \angle3\ form\ a\ linear\ pair&Adjacent angles on a straight line\\ \angle2\ is\ supplementary\ to\ \angle3&Deifinition of linear pair\\ m\angle2+m\angle3=180^o&Deifinition of supplementary \angle s\\ m\angle1+m\angle3=180^o&Substitution Property \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AStatement%26Reason%5C%5C%5B1ex%5D%0ALine%20m%20is%20parallel%20to%20line%20n%26Given%5C%5C%0A%5Cangle1%5Ccong%5Cangle2%26Corresponding%20angles%5C%5C%0Am%5Cangle1%3Dm%5Cangle2%26Deifinition%20of%20Congruent%20angles%5C%5C%0A%5Cangle2%5C%20and%5C%20%5Cangle3%5C%20form%5C%20a%5C%20linear%5C%20pair%26Adjacent%20angles%20on%20a%20straight%20line%5C%5C%0A%5Cangle2%5C%20is%5C%20supplementary%5C%20to%5C%20%5Cangle3%26Deifinition%20of%20linear%20pair%5C%5C%0Am%5Cangle2%2Bm%5Cangle3%3D180%5Eo%26Deifinition%20of%20supplementary%20%5Cangle%20s%5C%5C%0Am%5Cangle1%2Bm%5Cangle3%3D180%5Eo%26Substitution%20Property%0A%5Cend%7Btabular%7D)
</span>
Answer:
12
Step-by-step explanation:
4/5 x -6 = -2
4 x -30 = -10
-120 = -10
12
Answer: Choice D)
F(x) > 0 over the inverval (-infinity, -4)
Translation: The y or f(x) values are positive whenever x < -4.
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Further Explanation:
Recall that y = f(x), so if we say something like f(x) < 0 then we mean y < 0. Choice A is false because points on the curve to the left of x = -4 have positive y coordinates. Similar reasoning applies to choice B as well.
Choice C is false because while the interval (-infinity, -4) is above the x axis, the portion from x = -4 to x = -3 is below the x axis.
Choice D is true because everything to the left of x = -4 is above the x axis. Pick any point on the blue curve that is to the left of x = -4. This point will be above the horizontal x axis. Keep in mind that the parenthesis notation attached to the -4 means we dont include -4 as part of the interval.
