If you know the formular a^3+b^3=(a+b)(a^2-ab+b^2), you can solve this problem.
8 is 2 cubed, so x^3+2^3=(x+2)(x^2-2x+4)
so the other quadratic factor is x^2-2x+4
The equation we will use here is A^2+B^2=C^2, which is also know as the Pythagorean Theorem.
The given values are 6 and 9, where they can represent any value, there true values in the equation would be 36(6), and 81(9), so you must select a value that makes the equation true, given the constraints.
with that being said 3, doesnt work because
·36(6)+9(3)≠81(9)
·9(3)+81(9)≠36(6)
·36(6)+81(9)≠9(3)
10 doesnt work either because
·36(6)+81(9)≠100(10)
·81(9)+100(10)≠36(6)
·100(10)+36(6)≠81(9)
12 doesnt work either because
·144(12)+36(6)≠81(9)
·36(6)+81(6)≠144(12)
·81(9)+144(12)≠36(6)
If you see where this is going you would know that there is no valid solution here, however rounding is always a possibility, when you actually do the math 81(9)+36(6)=117, and when squared you get your answer of 10.8, and the closest answer is 10, there fore your answer would be 10
-I hope this is the answer you are looking for, feel free to post your questions on brainly at any time.
Answer:
.0625 or A
Step-by-step explanation:
You have to work out the mean. (The simple average of those numbers) Then for each number subtract the Mean and square the result. Then work out the average of those squared differences.
Hello,
f(x)-f(a)= -3x²-5x+1-(-3a²-5a+1)=-3(x²-a²)-5(x-a)=-3(x-a)(x+a)-5(x-a)
=-(x-a)(3(x+a)+5)
=-(x-a)(3x+3a+5)
lim (f(x)-f(a))/(x-a)=- lim (3x+3a+5)=3a+3a+5=-6a-5
if a=1==>-6*1-5=-11
Otherwise
f'(x)=-6x-5
f'(1)=-6-5=-11
at point(1,-7)
4a+2 because you divide 16a+8 by 4 because a square has 4 sides
