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BARSIC [14]
4 years ago
8

Round 479.8983 to the nearest whole number:

Mathematics
2 answers:
mixer [17]4 years ago
6 0
The answer would be 480 because the first number after the decimal is higher than 4. Hope this helps:)

~Ash
Slav-nsk [51]4 years ago
4 0
The answer to your question is 480

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Which of the following is the best definition of the domain
GalinKa [24]

Answer:

Step-by-step explanation:

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3 years ago
How are 25:5 and 10:2 equivalent ratios?
stira [4]

Answer:

Multipy by 2.5

Step-by-step explanation:

If you multiply 10 and 2 by 2.5 you would get 25 and 5

7 0
3 years ago
An amusement park charges and admission fee of 30 dollars for each person. Let C be the cost (in dollars) of admission for P peo
bogdanovich [222]

Answer:

14

Step-by-step explanation: B is the midpoint of AC, in other words it is the halfway point.

So A to B should be equal to B to C

Our expression is:

2x + 9 = 37

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5 0
3 years ago
Identify the asymptotes and state the end behavior of the function f(x)=5x/x-25
AnnyKZ [126]

Using it's concepts, it is found that for the function f(x) = \frac{5x}{x - 25}:

  • The vertical asymptote of the function is x = 25.
  • The horizontal asymptote is y = 5. Hence the end behavior is that y \rightarrow 5 when x \rightarrow \infty.

<h3>What are the asymptotes of a function f(x)?</h3>

  • The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
  • The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity. They also give the end behavior of a function.

In this problem, the function is:

f(x) = \frac{5x}{x - 25}

For the vertical asymptote, it is given by:

x - 25 = 0 -> x = 25.

The horizontal asymptote is given by:

y = \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{5x}{x - 25} = \lim_{x \rightarrow \infty} \frac{5x}{x} = 5

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

8 0
2 years ago
Derive the equation of the parabola with a focus at (3,1) and a directrix of y = 5
serg [7]
So hmmm  check the picture below

so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both

in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3

now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

because the parabola opens downwards, "p" is negative, and thus, -2

now, let's plug all those fellows in then

\bf \begin{array}{llll}&#10;(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;&#10;\begin{cases}&#10;h=3\\&#10;k=3\\&#10;p=-2&#10;\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)&#10;\\\\\\&#10;-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}

5 0
4 years ago
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