What is the relationship between the numerator and the denominator in a fraction?

Determine whether the relation is a function. {(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}

Gennadij [26K]

Answer:

The relation is a function.

Step-by-step explanation:

In order for the relation to be a function, every input must only have one output. Basically, you can't have 2 outputs for 1 input but you can have 2 inputs for 1 output. Looking at all of the points in the relation, we see that no input has multiple outputs, so the answer is yes, the relation is a function.

7 0

3 years ago

What is the equation of this line in standard form? −8x+7y=25 −9x+8y=−23 −8x+9y=−23 −8x+9y=23

kvasek [131]

−8x+7y=25 is already in standard form, altho some people would prefer to re-write it as

−8x+7y-25 = 0.

You have shared 4 equations here. Next time, please separate them with commas or semi colons, or type just 1 equation per line, for increased clarity. Thanks.

5 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

3 years ago

You saved $300 to spend over the summer. You decide to budget $50 to spend each week.

Oduvanchick [21]

Answer:

Part c)

The Slope of the line is: m=-50 and represents the amount of money spent per week.

Part d)

The y-intercept is: c=300 and represents the maximum money we have that can be spend over the weeks (i.e. our maximum budget alowed).

Step-by-step explanation:

To solve this question we shall look at linear equations of the simplest form reading:

$y = mx+c$ Eqn(1).

where:

$y$ : is our dependent variable that changes as a function of x

$x$ : is our independent variable that 'controls' our equation of y

$m$ : is the slope of the line

$c$ : is our y-intercept assuming an $x$ ⇔ $y$ relationship graph.

This means that as $x$ changes so does $y$ as a result.

Given Information:

Here we know that $300 is our Total budget and thus our maximum value (of money) we can spend, so with respect to Eqn (1) here:

$c=300$

The budget of $50 here denotes the slope of the line, thus how much money is spend per week, so with respect to Eqn (1) here:

$m=50$

So finally we have the following linear equation of:

$y= - 50x + 300$ Eqn(2).

Notice here our negative sign on the slope of the line. This is simply because as the weeks pass by, we spend money therefore our original total of $300 will be decreasing by $50 per week.

So with respect to Eqn(2), and different weeks thus various $x$ values we have:

Week 1: $x=1$ we have $y= -50 *1 + 300 = -50 +300 = 250$ dollars.