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2 years ago

Which function represents a proportional relationship? A) f(x) = 3x 4 B) f(x) = 3 4x C) f(x) = 3x 4 + 3 4 D) f(x) = 3 4x + 3 4

Mathematics

2 answers:

mr_godi [17]2 years ago

7 0

Answer:b

Step-by-step explanation:

densk [106]2 years ago

6 0

Answer: OPTION B

Step-by-step explanation:

The equation of a line that passes through the origin is the following:

$y=mx$

Where "m" is the slope of the line.

By definition, Proportional relationships have the following form:

$y=kx$

Where "k" is the Constant of proportionality.

Then, as you can notice, if you graph a Proportional relationship you will get a line that passes through the origin.

Therefore, based on the explained above, you can conclude that the option that shows a function that represents a Proportional relationship is the Option B.

Then, this is:

$f(x)=\frac{3}{4}x$

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Give one combination of two factors whose product is 0.24, if one of the factors is a whole number greater than 1.

Nutka1998 [239]

0.24 is an even number....so 2 will be a factor

2x = 0.24
x = 0.24 / 2
x = 0.12

so 2 factors would be 2 and 0.12.....because 2 * 0.12 = 0.24

6 0

3 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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2 years ago

The student council at Livingston High School is making T-shirts to sell for a fundraiser, at a price of $16 apiece. The costs,

Allushta [10]

Answer:

5 shirts must be sold.

Step-by-step explanation:

First, set up the equation. Each shirt (x) is sold for $16:

For every amount (y) for x: (y)(x) = (16)(y)

The cost is $4, with the setup fee as $60.

(16)(y) = (4)(y) + 60

16y = 4y + 60

Isolate the variable (y). Note the equal sign, what you do to one side, you do to the other. First, subtract 4y from both sides.

16y (-4y) = 4y (-4y) + 60

16y - 4y = 60

12y = 60

Next, isolate the variable (y). Divide 12 from both sides.

(12y)/12 = (60)/12

y = 60/12

y = 5

5 shirts must be sold.

To check, plug in 5 for y in the equation:

16y = 4y + 60

16(5) = 4(5) + 60

Simplify.

16(5) = 4(5) + 60

80 = 4(5) + 60

80 = 20 + 60

80 = 80 (True).

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2 years ago

I need some math help

Amiraneli [1.4K]

i think the answer is(1,0) im not sure

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3 years ago

5 1/3 + ( 2 1/3 + 1 1/3)=

Fiesta28 [93]

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3 years ago

Read 2 more answers