-- Carefully cut <em>each bar into 6 equal pieces</em>.
There are (5 x 6) = 30 pieces all together.
Each piece is 1/6 of a bar.
-- Each child gets 5 of the pieces.
That's the same as <em>5/6 of a whole bar</em>.
(6 children) x (5 pieces) = 30 pieces. It works out. yay !
The answer is (-2, 10) because when reflecting over the x, change the sign of y. When reflecting over the y, change the sign of x.
Answer:
(2x^2+5) x (x-2)
Step-by-step explanation:
Answer:
1/63
Step-by-step explanation:
Here is the complete question
In an experiment, the probability that event A occurs is 1
/7 and the probability that event B occurs is 1
/9
.
If A and B are independent events, what is the probability that A and B both occur?
Simplify any fractions.
Solution
the probability of independent events A and B occurring is P(A u B) = P(A)×P(B) where P(A) = probability that event A occurs = 1
/7 and P(B) = probability that event B occurs = 1
/9
.
So, P(A u B) = P(A)×P(B) = 1/7 × 1/9 = 1/63
Answer:
Hello your question is incomplete below is the complete question
What is wrong with the equation? integral^2 _3 x^-3 dx = x^-2/-2]^2 _3 = -5/72 f(x) = x^-3 is continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous at x = -3, so FTC2 cannot be applied The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. If f(2) = 14, f' is continuous, and f'(x) dx = 15, what is the value of f(7)? F(7) =
answer : The value of f(7) = 29
Step-by-step explanation:
Attached below is the detailed solution
Hence : F(7) - 14 = 15
F(7) = 15 + 14 = 29
