(4y 2 – 5y) + (3y – 7y 2) – (2y 2 + 6y – 5) Simplify.
1 answer:
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_______________
• Function: f(x) = 3x + 12.
A. Finding the inverse of f.
The composition of f with its inverse results in the identity function:
(f o g)(x) = x
f[ g(x) ] = x
3 · g(x) + 12 = x
3 · g(x) = x – 12
x – 12
g(x) = ⸺⸺
3
x
g(x) = ⸺ – 4 <——— this is the inverse of f.
3
________
B. Verifying that the composition of f and g gives us the identity function:
•
![\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Df%5Cbig%5Bg%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D3%5Ccdot%20%5Cleft%28%5Cdfrac%7Bx%7D%7B3%7D-4%5Cright%29%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%5Cdfrac%7Bx%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-3%5Ccdot%204%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx-12%2B12%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
and also
•
![\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Dg%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7Bf%28x%29%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7B3x%2B12%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7B%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%28x%2B4%29%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%2B4-4%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
________
C. Since f and g are inverse, then
f(g(– 2))
= (f o g)(– 2)
= – 2 <span>✔
</span>
• Call h the compositon of f and g. So,
h(x) = (f o g)(x)
h(x) = x
As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).
I hope this helps. =)
_______________
• Function: f(x) = 3x + 12.
A. Finding the inverse of f.
The composition of f with its inverse results in the identity function:
(f o g)(x) = x
f[ g(x) ] = x
3 · g(x) + 12 = x
3 · g(x) = x – 12
x – 12
g(x) = ⸺⸺
3
x
g(x) = ⸺ – 4 <——— this is the inverse of f.
3
________
B. Verifying that the composition of f and g gives us the identity function:
•
and also
•
________
C. Since f and g are inverse, then
f(g(– 2))
= (f o g)(– 2)
= – 2 <span>✔
</span>
• Call h the compositon of f and g. So,
h(x) = (f o g)(x)
h(x) = x
As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).
I hope this helps. =)
Answer:
The Recursive formula for the sequence is:
aₙ = aₙ₋₁ + d
The Explicit formula for the sequence is:
Step-by-step explanation:
Given the sequence
3,7,11,15
Here:
a₁ = 3
computing the differences of all the adjacent terms
7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4
The difference between all the adjacent terms is the same and equal to
d = 4
We know that a recursive formula basically defines each term of a sequence using the previous term(s).
The recursive formula of the Arithmetic sequence always involves the first term.
a₁ = 3
We know that, in the Arithmetic sequence, every next term can be obtained by adding the common difference and the preceding term.
so
The recursive formula of the sequence is:
aₙ = aₙ₋₁ + d
substitute n = 2 to find the 2nd term
a₂ = a₂₋₁ + d
a₂ = a₁+ d
substitute a₁ = 3 and d = 4
a₂ = 3 + 4
a₂ = 7
Thus, the recursive formula for the sequence 3,7,11,15 is:
aₙ = aₙ₋₁ + d
<u>An explicit rule for the sequence</u>
Given the sequence
3,7,11,15
We already know that
a₁ = 3
d = 4
An arithmetic sequence has a constant difference 'd' and is defined by
substituting a₁ = 3 and d = 4
Therefore, an explicit rule for the sequence
Answer:
This data suggest that there is more variability in low-dose weight gains than in control weight gains.
Step-by-step explanation:
Let be the variance for the population of weight gains for rats given a low dose, and
the variance for the population of weight gains for control rats whose diet did not include the insecticide.
We want to test vs
. We have that the sample standard deviation for
female control rats was
g and for
female low-dose rats was
g. So, we have observed the value
which comes from a F distribution with
degrees of freedom (numerator) and
degrees of freedom (denominator).
As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.