X² <span>+ 11x + 7
because 7 is a prime number, this doesn't factor prettily. you'll want to use the quadratic formula; if you aren't familiar with it, i'd either research it or look it up in your textbook, because it's clunky and not easily understood in this format:
(-b </span>± √((b)² - 4ac))/(2a)
in your equation x² + 11x + 7 ... a = 1, b = 11, and c = 7. what you do is you take the coefficients of every term, then plug it into your equation:
(-11 ± √((11)² - 4(1)(7))/(2(1))
not pretty, i know. but, regardless, you can simplify it:
(-11 ± √((11)² - 4(1)(7))/(2(1))
(-11 ± √(121 - 28))/2
(-11 ± √93)/2
and you can't simplify it further. -11 isn't divisible by 2, and 93 doesn't have a perfect square that you can take out from beneath the radical. the ± plus/minus symbol indicates that you have 2 answers, so you can write them out separately:
(x - (-11 - √93)/2) and (x + (-11 - √93)/2)
they look confusing, but those are your two factors. they can be simplified just slightly by changing the signs in the middle due to the -11:
(x + (11 + √93)/2) (x - (11 - √93)/2)
and how these would read, just in case the formatting is too confusing for you: x plus the fraction 11 + root 93 divided by 2. the 11s and root 93s are your numerator, 2s are your denominator.
The answer will be f √7 / 3 + 5 √7 / 14.
Every x coordinate of the first graph would be a multiple of 5. Example when x is 2, y = 8. The (x,y) coordinates for the first graph becomes (10,8). The second graph has a y value of 0 when x is 2. So its coordinate becomes (2,0). We see a trend here, the x coordinate of the second graph is 1/5 the x coordinate of the first graph.
I needed help on the same one have you gotten help or the answer?
Factor and find commmon ones
25=1*5*5
33=1*3*11
GCF=1
(I am noticing that GCF=1 for many of your questions)
