Parallel: the lines have = slopes. We thus need 5/6 to equal 2/p. then 5p must equal 12, and p = 12/5. (answer)
Check: Is 5/6 = 2 / (12/5)? YES
Perp.: The lines have slopes that are negative reciprocals of one another.
Then -6/5 = 2/p, or
-6 2
---- = ----
5 p Thus, -6p = 10, and p = -5/3 (answer)
Answer:
there are 11 big-bags and 24 small-bags.
Step-by-step explanation:
there are x big-bags and y small-bags.
so now we can know:
(1) x + y = 35
(2) 12x + 7y = 300
in (1),we can do like this:
both left and right x7
then (x + y)x7 = 35 x 7
then 7x + 7y = 35 x 7
then 7x + 7y = 245
now,
(1)7x + 7y =245
(2)12x + 7y = 300
we can both left and right do this: (2) - (1)
then
(12x + 7y) - (7x + 7y) = 300 -245
then
12x + 7y - 7x - 7y = 300 -245
then
12x - 7x +7y - 7y =300 -245
then
5x =55
then
5x ÷ 5 = 55 ÷ 5
then
x = 11
because
x + y =35;x=11
so
11 + y =35
11+ y -11 = 35 -11
then
y = 24
now we know:there are 11 big-bags and 24 small-bags.
Answer:
y < 10 or (10, -∞)
Step-by-step explanation:
7y + 1 > 8y - 9
Simply get the y's on the right by subtracting 7y from both sides:
1 > y - 9
then adding 9 to both sides:
y < 10
or in interval notation:
(10, -∞)
If 1 and 2 becomes multiplayed gave 3 and 3+3=4 (4-2)+(4-1)= 2+3=4 and 4 is n.
Notation
I imagine that the expression you are asked to work with is:

When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:

To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,

. For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets.
![[( x^{3}+5x)]-[3 x^{2} -15]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D)
Let's next factor what is in each bracket:
![[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D%20%3D%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D)
Notice that both brackets have the same expression in them so now we factor that out:
![[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)](https://tex.z-dn.net/?f=%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D%20%3D%20%28x-3%29%28%20x%5E%7B2%7D%20%2B5%29)
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: