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3 years ago

Suppose s varies directly with t. When t=27, s=3. What is s when t=63?

Mathematics

2 answers:

yan [13]3 years ago

5 0

Answer:

S=7

Step-by-step explanation:

sorry if this is wrong

adell [148]3 years ago

3 0

Answer: s = 7

Step-by-step explanation:

27 is simply 9 times 3.

Thus, simply do 63/9 to get 7.

Hope it helps :)

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Point V is located at -16. Points W and X are each 7 units away from Point V. Where are W and X located?

nasty-shy [4]

Answer:

Location of W is - 23, location of X is - 9.

Step-by-step explanation:

location of V = - 16

Points W and X are each 7 units away from Point V.

Let the W is at left of V and X is right of V.

location of W = -16 - 7 = - 23

location of X = - 16 + 7 = - 9

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3 years ago

What type of symmetry does this figure have?

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B. Is the answer because you can fold it in half

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3 years ago

Please help with solving this problem....!!!

kifflom [539]

Answer:

The Average Temperature for the week is -1°C

Step-by-step explanation:

First to find the average, you have to combine all the numbers

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4 years ago

A group of learners is trying to identify the vertices of the feasible region from the graph shown below during a Live Classroom

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Answer:

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Step-by-step explanation:

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4 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.)

6 0

3 years ago