Answer:
6x^2 - 3x

Step-by-step explanation:
3x (2x-1)
multiply 3x by 2x -> 6x^2
multiply 3x by -1 -> -3x
You had the right idea using the Pythagorean theorem to solve for b.
Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.
Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.
5² + b² = (2√2)²
25 + b² = 2²(√2)²
25 + b² = 4(2)
25 + b² = 8
b² = 8 - 25
b² = - 17
b = √-17
b= (√17i)
Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.
Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,


which is a cubic polynomial in
with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,

where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then
can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)
Answer:
A
<em>brainliest is much appreciated!</em>
Step-by-step explanation:
hope this helps!
Since 800 is greater than 0.8 your statement would be 800>0.8