Frankie entered a sweepstakes to win a grand prize of $10,500,000,00. Express this amount using scientific notation

skad [1K]

You have to add 3+3+5.
It equals 11.
Then divide 132 by 11.
It equals 12.
Then you have to put it into the ratio, so you multiply each number by 12.
So the ratio is 36:36:60

3 0

3 years ago

True or false: The y-intercept is (0.3) and it represents the initial cost of the ride.

Leni [432]

The answer is True.

Since we are talking about the cost of something, and whenever you have a y-intercept, it means the price.

The y-intercept represents the initial cost of the taxi ride then how many miles you ride for, the more you pay plus the initial cost.

Best of Luck!

7 0

3 years ago

A factory produces 37,650 T shirts in a 30-day month.

yan [13]

1,255 a day at average

5 0

2 years ago

Read 2 more answers

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

2 years ago

Find all real and imaginary solutions to the equation x^3-4x^2+x-4=0

lawyer [7]

Step-by-step explanation:

STEP 1

Equation at the end of step 1

(((x3) - 22x2) + x) - 4 = 0

STEP 2

Checking for a perfect cube

2.1 x3-4x2+x-4 is not a perfect cube

5 0

2 years ago