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3 years ago

13

The function f(x) = is translated up 4 units. Which equation represents the translated function?

2 answers:

r-ruslan [8.4K]3 years ago

5 0

Answer:

f(x)= is translated up four units can be written as

f(x)+4

Step-by-step explanation:

kkurt [141]3 years ago

3 0

Answer:

The translated function can be written as:

y = f(x) + 4

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A regression analysis between sales (y in $1000) and advertising (x in dollars) resulted in the following equation y = 50,000 +

Vikentia [17]

Answer:

D. increase of $1 in advertising is associated with an increase of $6,000 in sales.

Step-by-step explanation:

Every dollar in advertisting is multiply by 6 in the equation, but every unit in y represent $1000. That means that $1 1n advertisting is multiply by 6 and then by $1000, that is $1 in advertisting = $6000 in sales.

4 0

3 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

3 years ago

I need help with number 18

algol13

Answer:

61.6666666667 =61.67

Step-by-step explanation:

4 0

3 years ago

30 ounces at $4.00 per pound. how much was paid for 30 ounces

Nata [24]

It would be $7.50 because 4 ounces equals $1 so 30/4 =7.5

3 0

3 years ago

Tpy6a [65]

Answer:

angle abc=125°

.......................

6 0

3 years ago

Read 2 more answers