10. Since it you want to find how many 0.3s it would take to make 3, you would divide 3 by 0.3 or move the decimal over one spot to the right making it 10.
Answer:
-3.98 (nearest hundredth)
Step-by-step explanation:
The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:
Given interval: -2 ≤ x ≤ 2
Substituting the values into the equation:
Let's solve this problem step-by-step.
STEP-BY-STEP EXPLANATION:
Let's first establish that triangle BCD is a right-angle triangle.
Therefore, we can use Pythagoras theorem to find BC and solve this problem. Pythagoras theorem is displayed below:
a^2 + b^2 = c^2
Where c = hypotenus of right-angle triangle
Where a and c = other two sides of triangle
Now we can solve the problem by substituting the values from the problem into the Pythagoras theorem as displayed below:
Let a = BC
b = DC = 24
c = DB = 26
a^2 + b^2 = c^2
a^2 + 24^2 = 26^2
a^2 = 26^2 - 24^2
a = square root of ( 26^2 - 24^2 )
a = square root of ( 676 - 576 )
a = square root of ( 100 )
a = 10
Therefore, as a = BC, BC = 10.
If we want to check our answer, we can substitute the value of ( a ) from our answer in conjunction with the values given in the problem into the Pythagoras theorem. If the left-hand side is equivalent to the right-hand side, then the answer must be correct as displayed below:
a = BC = 10
b = DC = 24
c = DB = 26
a^2 + b^2 = c^2
10^2 + 24^2 = 26^2
100 + 576 = 676
676 = 676
FINAL ANSWER:
Therefore, BC is equivalent to 10.
Please mark as brainliest if you found this helpful! :)
Thank you and have a lovely day! <3
It is A, 1/5 hopefully i am the BRAINLIEST!!!!!
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, ∞)