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3 years ago

Which situation represents a non-proportional relationship?

Mathematics

1 answer:

denis-greek [22]3 years ago

6 0

Answer:

Correct option:

"The total cost of renting a jet ski for $25 per hour plus a fee of $50."

Step-by-step explanation:

Proportional relationships are associations between two variables where the ratio of these two variables are correspondent. Or, in a proportional relationship, one variable is a constant value times the other variable. This constant is known as the "constant of proportionality".

The first three options represents a proportional relationship.

But the last option does not.

"The total cost of renting a jet ski for $25 per hour plus a fee of $50."

That is,

TC = 25·h + 50

The relationship is linear not proportional.

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The product of two numbers

sladkih [1.3K]

Answer:

The product of two numbers is the result you get when you multiply them together. So 12 is the product of 3 and 4, 0 is the product of 4 and 5 and so on.

Step-by-step explanation:

4 0

3 years ago

Find the slope(6,8) and (9,10)

USPshnik [31]

name the points

a=(x1,y1) b=(x2,y2)

a=(6,8) b=(9,10)

use the slope formula

$m=\frac{y2-y1}{x2-x1}$

replace

$\begin{gathered} m=\frac{10-8}{9-6} \\ m=\frac{2}{3} \end{gathered}$

answer= The slope is equal to 2/3

a=(9,10) b=(6,8)

using the formula

$\begin{gathered} m=\frac{y2-y1}{x2-x1} \\ m=\frac{8-10}{6-9} \\ m=\frac{-2}{-3}=\frac{2}{3} \end{gathered}$

slope will also be 2/3

5 0

1 year ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$