It’s c) XS, i love your pfp btw!! :)
48 all you had to do was out it in a calculator that's what I did
Answer:
Step-by-step explanation:
The answer is 26
Step 1: move all terms to one side
7 + 2x^2 - 10x = 0
Step 2: Use the quadratic formula
X = 10 + 2(square root of 11). 10-2 _/—-11
————————————. ——————
4. , 4
Step 3: Simplify solutions
X= 5 + (square root of 11) 5 - (square root 11)
——————————- , ————————
2. 2
Answer:
The first set of consecutive even integers equals (8 , 6)
The second set is ( - 8 and - 6) which also works.
Step-by-step explanation:
Equation
(x)^2 + (x + 2)^2 = (x)(x + 2) + 52 Remove the brackets on both sides
Solution
x^2 + x^2 + 4x + 4 = x^2 + 2x + 52 Collect the like terms on the left
2x^2+ 4x+ 4 = x^2 + 2x + 52 Subtract right side from left
2x^2 - x^2 + 4x - 2x + 4 - 52 = 0 Collect the like terms
x^2 + 2x - 48 = 0 Factor
(x + 8)(x - 6) = 0
Answer
Try the one you know works.
x - 6 = 0
x = 6
Therefore the two integers are 6 and 8
6^2 + 8^2 = 100
6*8 + 52 = 100
So 6 and 8 is one set of consecutive even numbers that works.
========================
What about the other set.
x + 8 = 0
x = - 8
x and x + 2
- 8 and -8 + 2 = - 8, - 6
(- 8 )^2 + (- 6)^2 = 100
(-8)(-6) + 52 = 100
Both sets of consecutive numbers work.