Answer:
She buys 5 boxes of popcorn and 6 boxes of candy
Step-by-step explanation:
this is because 5 times 4.00 is 20$ and 2.50 times 6 is 15$
15 + 20 is 35
so your answer is 5 boxes of popcorn and 6 boxes of candy
hope this helped :D
1) False
Adjacent angles must share a common side/ray.
2) B and D
Adjacent angles are those that are directly next to each other and share a common side.
3) C
Angles 1 and 2 are congruent. Angles 3 and 4 are congruent. Both of these pairs have angles that are opposite each other.
4) B
Angles 1 and 2 add up to 180. Angles 3 and 4 add up to 180. Both of these pairs of angles are supplementary.
5) A
These angles are directly next to each other and share a common side.
Hope this helps!! :)
Answer: x=12, x=-3
Step-by-step explanation: For this factored problem, you see that -36 is negative, as well at -9. This shows you that you need to find two numbers that multiply to get -36, and add up to get -9. Two numbers that both satisfy these requirements are -12 and 3. They multiply to get -36, and add to get -9.
Once you have found your values, you plug them into (x+/- __)(x+/- __), which is (x-12)(x+3).
But since the problem asks you to find the zeroes, you set each factor equal to zero, like this:
x-12=0 and x+3=0
Then solve and you have you answer!
if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.
so, any when we're referring to even roots like ![\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B2%5D%7B~~%7D%2C%5Csqrt%5B4%5D%7B~~%7D%2C%5Csqrt%5B6%5D%7B~~%7D....)
, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.
now, that's is not true for odd roots like ![\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B~~%7D%2C%5Csqrt%5B5%5D%7B~~%7D%2C%5Csqrt%5B7%5D%7B~~%7D....)
, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.
![\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B27%7D%5Cimplies%20%5Csqrt%5B3%5D%7B3%5E3%7D%5Cimplies%203%5Cqquad%20because%5Cqquad%20%283%29%283%29%283%29%3D27%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%28-3%29%28-3%29%28-3%29%5Cne%2027~%5Chspace%7B8em%7D%28-3%29%28-3%29%28-3%29%3D-27%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B-125%7D%5Cimplies%20%5Csqrt%5B3%5D%7B-5%5E3%7D%5Cimplies%20-5%5Cqquad%20because%5Cqquad%20%28-5%29%28-5%29%28-5%29%3D-125%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%285%29%285%29%285%29%5Cne%20-125~%5Chspace%7B10em%7D%285%29%285%29%285%29%3D125)
so, when the root is an odd root, you will always get only one number that will produce the radicand.
