Something is 25% off I write that as 25/100 How would i write what the real cost is to the cost with the percent off.

Which function has a greater rate of change? Function A: x 1.5 2.5 3.5 4.5 5.5 y 5 0 –5 –10 –15 Function B: y = –4.5x + 15 Funct

expeople1 [14]

Answer:

Function A has a rate of change of -5 and Function B has a rate of change of -4.5, so Function B has a greater rate of

Step-by-step explanation:

Function A:

$\begin{array}{cccccc}x & {1.5} & {2.5} & {3.5} & {4.5} & {5.5} \ \\ y & {5} & {0} & {-5} & {-10} & {-15} \ \end{array}$

Function B:

$y = -4.5x + 15$

Required

Which has a greater rate of change

The rate of change of function A is calculated as thus:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$(x_1,y_1) = (1.5,5)$

$(x_2,y_2) = (2.5,0)$

So, we have:

$m = \frac{0-5}{2.5 - 1.5}$

$m = \frac{-5}{1}$

$m = -5$

For function B

The general function is:

$y = mx + b$

Where

m is the rate of change

By comparing $y = mx + b$ to $y = -4.5x + 15$

$m = -4.5$

So, we have that:

Function A has a rate of -5; function B has a rate of -4.5.

By comparison: -4.5 is greater than 5

Hence, function B has a greater rate.

4 0

3 years ago

Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

4 years ago

What is the standard form of -9+8x=-10y

Nitella [24]

8x+10y=9

Hope this helps.

7 0

3 years ago

Read 2 more answers

How to solve 2v 14=20

LuckyWell [14K]

2v+14=20
2v=20-14
2v=6
V=3

8 0

3 years ago

Mr. Jenkins is going to buy museum tickets for a class trip. The museum director gives him the following prices:

ra1l [238]

For answer a the price will be the same because it's $125 for every 10 tickets

5 0

4 years ago