Answer:
<em>The solution is too long. So, I included them in the explanation</em>
Step-by-step explanation:
This question has missing details. However, I've corrected each question before solving them
Required: Determine the inverse
1:
Replace f(x) with y
Swap y & x
Divide through by 25
Replace y with f'(x)
2.
Replace g(x) with y
Swap y & x
Add 1 to both sides
Make y the subject
3:
Replace h(x) with y
Swap y & x
Add to both sides
Multiply through by -4
Divide through by 9
4:
Replace f(x) with y
Swap y with x
Take 9th root
Replace y with f'(x)
5:
Replace f(a) with y
Swap a with y
Subtract 8
Take cube root
Replace y with f'(a)
6:
Replace g(a) with y
Swap positions of y and a
Solve using quadratic formula:
; ;
becomes
Factorize
7:
Replace f(b) with y
Swap y and b
Open Brackets
Solve using quadratic formula:
; ;
becomes
Factorize:
Replace y with f'(b)
8:
Replace h(x) with y
Swap x and y
Cross Multiply
Subtract x from both sides:
Subtract 2y from both sides
Factorize:
Make y the subject
Replace y with h'(x)
9:
Replace h(c) with y
Swap positions of y and c
Square both sides
Subtract 2 from both sides
Make y the subject
10:
Replace f(x) with y
Swap positions of x and y
Cross Multiply
Subtract y from both sides
Add x to both sides
Factorize
Make y the subject
Replace y with f'(x)
Answer:
your answer is -2x-13y^2
Step-by-step explanation:
Probability is the likelihood or chance that an event will occur. The probability of P(AUB) is 1/2
<h3>Conditional probability</h3>
Probability is the likelihood or chance that an event will occur. Given the following parameters
If P(A) = 1/6
P(B) = 5/12
P(A\B) + P(B\A) = 7/10
Required
p(AUB)
Recall that:
P(A|B)=P(AnB)/P(B)
P(B|A) = P(BnA)/P(A)
P(AnB)/P(B) + P(BnA)/P(A) = 7/10
12/5P(AnB) + 6P(BnA) = 7/10
42/5P(BnA) = 7/10
6/5P(BnA) = 1/10
6P(BnA) = 1/2
P(BnA) = 1/12
<u>Determine P(AUB)</u>
P(AUB) = P(A) + P(B) - P(AnB)
P(AUB) = 1/6 + 5/12 - 1/12
P(AUB) = 1/6 + 4/12
P(AUB) = 2+4/12
P(AUB) = 1/2
Hence the probability of P(AUB) is 1/2
Learn more on probability here: brainly.com/question/24756209
Answer:
No
Step-by-step explanation: If right press thanks and rate 5/5 plz
Let U = {all integers}.
Consider the following sets:
A = {x | x ∈ U and x > 3}
B = {x | x ∈ U and x is an even integer}
C = {x | x ∈ U and 2x is an odd integer}
D = {x | x ∈ U and x is an odd integer}
<span>
Use the defined sets to answer the questions.
Assuming 0 is an even integer, which set is the complement to set B?
<span>
D
</span> Which set is an empty set?
<span>
C</span>
Which set would contain the subset E = {1, 3, 5, 7}?
<span><span>
D</span></span></span>