Hello! My work for this problem is attached below. The answer is
Vb = 4Va
Answer:
Step-by-step explanation:
what are the vocabulary words that u are supposed to use?
Answer:
if there are answer choices i would put the closest one to 16
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Answer:
w = 
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
w² + 7² = 14²
w² + 49 = 196 ( subtract 49 from both sides )
w² = 147 ( take the square root of both sides )
w =
