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BaLLatris [955]
3 years ago
12

the cost in dollars of a school banquet is 64 + 12n,where n is the number of people attending. what is the cost for 80 people.

Mathematics
2 answers:
nataly862011 [7]3 years ago
8 0
The total cost for 80 people attending the banquet is $1,024.00

If n represents the number of people, and 80 people are attending, plug in 80 for n:
64+12n \\ 64+12(80) \\ 64+960=1,024 dollars
koban [17]3 years ago
4 0
If n = the number of people attending, jut plug in 80 for n:

64+12n\\64+12(80)\\64+960\\1024

The cost for 80 people is $1024.
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Two trains leave two different stations 300 km apart; the first starts at noon and the second at 12 15 h. Travelling on parallel
alexgriva [62]

Answer:

See below

Step-by-step explanation:

From noon to 1500  is three hours

  train A    then travels  <u> 3x</u>   where x = speed in km/hr

  train B   only travels for 2.45 hours and covers

                  <u>   2.45 ( x + 15)  </u>        where x + 15 is its speed in km/hr

       these two values sum to the 300 km distance

3x     +     2.45 ( x + 15) = 300

3x + 2.45 x  + 36.75 = 300

x = 48.3 km/hr       x+15= 63.3 km / hr

8 0
2 years ago
Are 8x+ 16 and 4(2x+4) equivalent
8090 [49]
Yes they are equivalent
8 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

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