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scoundrel [369]

3 years ago

13

What transformation has changed the parent function f(x) = log4 x to its new appearance shown in the graph below?

1 answer:

torisob [31]3 years ago

4 0

We know the graph of the parent function will pass through the point (1, 0). If that is the point that is transformed, it is shifted 5 to the left and 5 up. The appropriate choice is

... f(x + 5) + 5

_____

The graph has an infinite number of points, any one of which can be transformed to the given point. There are an infinite number of transformations that will satisfy the problem requirements. However, only one of those listed will do so.

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Make sure to write be specific this is for my math test. i need help im as dumb as a rock

denis23 [38]

Answer:

y intercept is 420

Step-by-step explanation:

6 0

3 years ago

Fracciones equivalentes a 2/8

jeyben [28]

$\dfrac{2}{8} = \dfrac{2 \div 2}{8 \div 2} = \dfrac{1}{4}$

5 0

3 years ago

How much of an alloy that is 10% copper should be mixed with 600 ounce of an alloy that is 70%copper in order to get an alloy th

svetoff [14.1K]

Answer:

1200 ounces

Step-by-step explanation:

How much of an alloy that is 10% copper should be mixed with 600 ounce of an alloy that is 70%copper in order to get an alloy that is 30%copper?

Let us represent

Number of ounce of 10% copper = x

Hence, our Equation =

x × 10% + 600 × 70% = (x + 600) × 30%

0.1x + 420 = (x + 600)0.3

0.1x + 420 = 0.3x + 180

Collect like terms

0.3x - 0.1x = 420 - 180

0.2x = 240

x = 240/0.2

x = 1200 ounces

7 0

2 years ago

Who can help me d e f thanks

12345 [234]

d)

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

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.

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

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

e)

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

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.

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

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

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

f)

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

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.

7 0

1 year ago

I need help with these (need to show work too)

alexgriva [62]

Shut up and do you own work.

3 0

2 years ago