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3 years ago

15

If you apply the changes below to the absolute value parent function, f(x) = |x|, what is the equation of the new function?Shift

4 units to the right.Shift 6 units up.
A. g(x) = |x – 4| + 6

B. g(x) = |x – 6| + 4

C. g(x) = |x + 6| + 4

D. g(x) = |x + 4| + 6

2 answers:

grin007 [14]3 years ago

7 0

G(x)=|x-4|+6

answer: A

Regards M.Y.

ss7ja [257]3 years ago

6 0

Answer:

A. g(x) = |x – 4| + 6

Step-by-step explanation:

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3 years ago

Read 2 more answers

Expanding logarithmic Expression In Exercise,Use the properties of logarithms to rewrite the expression as a sum,difference,or m

ch4aika [34]

Answer:

$\ln x+\frac{1}{3}\ln (x^2+1)$

Step-by-step explanation:

Consider the given expression is

$\ln (x\sqrt[3]{x^2+1})$

We need to rewrite the expression as a sum,difference,or multiple of logarithms.

$\ln (x(x^2+1)^{\frac{1}{3}})$ $[\because \sqrt[n]{x}=x^{\frac{1}{n}}]$

Using the properties of logarithm we get

$\ln x+\ln (x^2+1)^{\frac{1}{3}}$ $[\because \ln (ab)=\ln a+\ln b]$

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Therefore, the simplified form of the given expression is $\ln x+\frac{1}{3}\ln (x^2+1)$ .

6 0

3 years ago

Olenka [21]

Answer:

d = -1/3, 0

Step-by-step explanation:

Subtract the constant on the left, take the square root, and solve from there.

(6d +1)^2 + 12 = 13 . . . . given

(6d +1)^2 = 1 . . . . . . . . . .subtract 12

6d +1 = ±√1 . . . . . . . . . . take the square root

6d = -1 ±1 . . . . . . . . . . . .subtract 1

d = (-1 ±1)/6 . . . . . . . . . . divide by 6

d = -1/3, 0

_____

Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:

f(x) = (6x+1)^ +12 -13

We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.

4 0

3 years ago