Answer:
C. Infinitely many solutions.
Step-by-step explanation:
-4x - 7 + 10x = -7 + 6x
Combine like terms on the left side. Rearrange the right side.
6x - 7 = 6x - 7
Add 7 to both sides.
6x = 6x
Subtract 6x from both sides.
0 = 0
0 = 0 is a true statement.
Both sides are equal, so all real values of x make the equation true.
Answer: C. Infinitely many solutions.
![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20~%5Chfill%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cboxed%7Ba%5E6%2Bb%5E6%7D%5Cimplies%20a%5E%7B2%5Ccdot%203%7D%2Bb%5E%7B2%5Ccdot%203%7D%5Cimplies%20%28a%5E2%29%5E3%2B%28b%5E2%29%5E3%20%5C%5C%5B2em%5D%20%5Ba%5E2%2Bb%5E2%5D%20%5B%28a%5E2%29%5E2-a%5E2b%5E2%2B%28b%5E2%29%5E2%5D%5Cimplies%20%5Cboxed%7B%28a%5E2%2Bb%5E2%29%28a%5E4-a%5E2b%5E2%2Bb%5E4%29%7D)
about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.
5, 8, 13 are no dice, namely 5² + 8² ≠ 13
25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²
however, 5,12 and 13 are indeed a pythagorean triple
also is 39, 80, 89.
when looking for a pythagorean triple, recall that c² = a² + b².
so the longest leg is the sum of the square of the small ones.
so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.
1. 2x+3=4x+2
⇒ 4x-2x=3-2
2x=1
x=1/2
There is one solution.
2. 18p-10p=5+6/2
8p=5+3
8p=8
p=8:8
p=1
3. 6a/3=2a
2a-a= -2 -3
a= - 5
4. 5+r-2=9,
5-2+r=9
3+r=9
r=9 - 3
r=6
5. -0,9 m= 9
m= - 10
6. -2 (x+1/4)=5-1=4
x+1/4= -2
x= -2 - 1/4
x= -8/4 - 1/4= - 9/4
7. 6f+4f= 6 + 12
10 f = 18
f=18/10=1,8
8. 7n - 3n= 16
4n=16
n=16:4
n=4
9. 1/3m-5/6m= - 15 - 3
2/6 m-5/6=-18
- 3/6m=-18 ⇒ 1/2m=18
m=2*18=36
Answer:
D
Step-by-step explanation:
Cuz sqr root of 20 = 2 sqrt 5, -2 sqrt 5
Answer:
Andre has 4 pounds of Broccoli and 6 pounds of Zucchini.
Step-by-step explanation: