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3 years ago

Describe how to find the sale price of an item that has been discounted 25%.

Mathematics

2 answers:

Delicious77 [7]3 years ago

7 0

$x$ = initial price
$y$ = new price

$y$ = 0.75 $x$

Nadya [2.5K]3 years ago

4 0

You subtract the % that the sale is and take it away from 100%. You get 75% then take the original price and divide it by .75

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A jacket at a clothing store is being discounted by 10%. If the jacket was originally priced at x dollars, which of the followin

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2 years ago

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At the end of a snow storm, Audrey saw there was a lot of snow on her front lawn.

Montano1993 [528]

An equation that represents the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling is S = 10 - 2t

In this question, there was a depth of 10 inches of snow on the lawn when the storm ended and then it started a melting at a rate of 2 inches per hour.

We need to write an equation for S in terms of t, snow representing the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling.

So, the equation would be,

S = 10 - 2t

Therefore, an equation that represents the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling is S = 10 - 2t

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1 year ago

Find BC round to the nearest tenth

slega [8]

Answer:

6.7

Step-by-step explanation:

Here, we can apply cosine law, $c^2=a^2+b^2-2abcos\theta$

We the plug in values: $(CB)^2+6^2+5^2-2*6*5*cos74$

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2 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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3 years ago