Step-by-step explanation:
It came from nowhere. It makes no sense to add up the balance numbers. To illustrate, let's use a different example:
![\left[\begin{array}{cc}Spend&Balance\\100&400\\100&300\\100&200\\100&100\\100&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSpend%26Balance%5C%5C100%26400%5C%5C100%26300%5C%5C100%26200%5C%5C100%26100%5C%5C100%260%5Cend%7Barray%7D%5Cright%5D)
Adding up the money you spent, and you get $500. Add up the balances, and you get $1000. But why would you add the balances? The 300 in the second line is included in the 400 in the first line. You can't add them together. You'd be counting the 300 twice.
First, factor out a 3.
3(x² - 9)
In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.
Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
<u>3(x-3)(x+3)</u>
There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
Answer:
<u>The missing angle measures also 110°</u>
Step-by-step explanation:
Let's recall that If the transversal cuts across parallel lines (apparently, this particular case) then those alternate exterior angles have the same measure (.∠ 110° and ∠?) So in the figure given, the two alternate exterior angles measure the same.
<u>The missing angle measures also 110°</u>
Answer: 4^2/5 is your main answer and pls mark me as brainiest
Step-by-step explanation:
Answer:
-1 -1 -3 123456
Step-by-step explanation:
