Let us assume for starters that it is true.
The area of the square is s^2, so that is where you start.
(1 + 2*sqrt(2) ) ^2 = 1 + 4*sqrt(2) + 8 Taking 1 away form that does not leave you with 8.
Let's start again.
If all sides are reduced by 1 , but the square is retained, then the length of the side is 2sqrt(2). Now we have
s^2 = (2sqrt(2) )^2 = 4 * 2 = 8
The question is really not worded clearly enough to give an answer. I think it's true, but the language has to be twisted a bit.
Answer: I'd answer true, but don't be surprised if it is not interpreted the way I did on my second try.
Answer:
D) 11x + 9
Step-by-step explanation:
3(2x + 5) - (-5x + 4) = (6x + 15) - (-5x + 4) = 6x + 15 + 5x - 4
add like terms:
(6x + 5x) + (15 - 4) = 11x + 9
Answer: C
Step-by-step explanation:
yes
I believe the correct answer is C I could be wrong tho
Answer:
x-intercepts: -5 +/-
. y-intercept: 4
Step-by-step explanation:
For x-intercepts use the quadratic equation
![x= (-b +/-\sqrt{b^2-4ac} )/2a](https://tex.z-dn.net/?f=x%3D%20%28-b%20%2B%2F-%5Csqrt%7Bb%5E2-4ac%7D%20%29%2F2a)
Fill in your values:
![x= (-10 +/-\sqrt{100-16} )/2](https://tex.z-dn.net/?f=x%3D%20%28-10%20%2B%2F-%5Csqrt%7B100-16%7D%20%29%2F2)
And solve
![x=-5 +/- \sqrt{21}](https://tex.z-dn.net/?f=x%3D-5%20%2B%2F-%20%5Csqrt%7B21%7D)
For y-intercepts set x=0 and solve. In this case, setting x= 0 gives you y=4.