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4 years ago

Suppose that g(x) = f(x) + 3. which statement best compares the graph of g(x) with the graph of f(x)?

Mathematics

2 answers:

otez555 [7]4 years ago

8 0

Answer:

C. the graph of g(x) is shifted 3 units up.

Step-by-step explanation:

The statement that compares the graph of g(x) with the graph of f(x) is that the graph of g(x) is shifted 3 units up.

If we have the equation:

G(x) = F(x) + 3

Then, the graph of G(x) will be composed of points 3 units above the graph of F(x).

Conversely, the difference between F(x) and G(x) at any particular value of x will always be 3.

tatiyna4 years ago

6 0

Answer:

C. the graph of g(x) is shifted 3 units up

Step-by-step explanation:

Given: g(x) = f(x) + 3

Here the graph of g(x) = the graph of f(x) + 3, where 3 represents the vertical shift.

Since it is 3, so the graph moves 3 units up.

Answer: C. the graph of g(x) is shifted 3 units up.

Hope you this will helpful.

Thank you.

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3 years ago

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12345 [234]

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

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2 years ago

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ad-work [718]

Answer:

The answer is: Store A is .41 cents per bagel and store B is .45 cents per bagel.

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2.05/5 = .41 cents per bagel.

Store B is 3 bagels for $1.35.

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So store A is the least expensive one.

Hope this Helps! Have an Awesome Day!! (-:

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