If the triangle is a right triangle, the pythagorean theorem would work
since it doesn’t, the triangle is NOT A RIGHT TRIANGLE
(see attached pic for work :))
Hello from MrBillDoesMath!
Answer:
40
Discussion:
A diagram is always appreciated!
Assuming that
mAOC = mAOB + mBOC =>
108 = (3x + 4) + (8x - 28) => combine common terms
108 = (3x + 8x) + (4 - 28 ) =>
108 = 11x - 24 => add 24 to both sides
132 = 11x =>
x = 132/11 = 12
So mAOB = 3x + 4 = 3(12) + 4 = 36 + 4 = 40
Thank you,
MrB
Answer:
A) x = 3 or -1
B) x = -7
C)x = -7
Step-by-step explanation:
A) x² + 2x + 1 = 2x² - 2
Rearranging, we have;
2x² - x² - 2x - 2 - 1 = 0
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
B) ((x + 2)/3) - 2/15 = (x - 2)/5
Multiply through by 15 to get;
5(x + 2) - 2 = 3(x - 2)
5x + 10 - 2 = 3x - 6
5x - 3x = -6 - 10 + 2
2x = -14
x = -14/2
x = -7
C) log(2x + 3) = 2log x
From log derivations, 2 log x is same as log x²
Thus;
log(2x + 3) = logx²
Log will cancel out to give;
2x + 3 = x²
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
Answer:
23
Step-by-step explanation:
