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Lorico [155]
3 years ago
5

A school is having a fund raising dance and hope to raise at least $4,000 for the athletic department. There are two types of ti

ckets available for the dance, standard and VIP. The standard ticket costs $25 each, and the VIP tickets cost $70.
x = the number of standard tickets


y = the number of VIP tickets
Mathematics
1 answer:
azamat3 years ago
8 0

Answer:

25X + 70Y = 4000  or  Y= - 5x/14 + 40/7

Step-by-step explanation:

the combination of the X tickets and the Y tickets combined need to equal 4000

X and Y both stand for the amount of tickets that are all possible to occur to get the total of 4000.

the first formula represents the normal formula

the second formula is in y intercept form

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The first has value 2500*(1+.05*9)=3625.  The second, similarly, has value 4160, and the third, 3400.  Thus, the order, from least to greatest, is third, first, second.
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The sum of Pete's and Sam's ages is 30. Five years ago, Pete was 3 times as old as Sam. How old is Sam?
RUDIKE [14]
P-5=3s-15 because Pete is 20 and Sam is 10. 20-5= 15 and 3(10)-15 also equals 15. The answer is C.
4 0
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Which of the following represents a function?
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Answer:

1 i koow

Step-by-step explanation:

6 0
3 years ago
Expand the following:<br> a) x(x + 2)<br> b) x(2x - 5)<br> c) 2x(3x + 4)<br> d) 6x(x - 2)
TEA [102]

Answer:

<h2>a. \:  {x}^{2}  + 2</h2>

<h2> </h2><h2>b. \: 2 {x}^{2}  - 5x</h2><h2> </h2><h2>c. \: 6 {x}^{2}  + 8x</h2><h3> </h3><h3>d. \: 6 {x}^{2}  - 12x</h3>

solution,

a. \: x(x + 2) \\  \:  \:  = x \times x + 2 \times x \\  \:  \:  =  {x}^{2}  + 2x

b . \: x(2x - 5) \\  \:  = x \times 2x - x \times 5 \\  \:  \:  = 2 {x}^{2}  - 5x

c. \: 2x(3x + 4) \\  \:  = 2x \times 3x + 2x \times 4 \\  \:  \:  = 6 {x}^{2}  + 8x

d. \: 6x(x - 2) \\  \:  \:  = 6x \times x  - 6x \times 2 \\  \:  \:  = 6 {x}^{2}  - 12x

<h2> </h2><h2 />

Hope this helps...

Good luck on your assignment..

4 0
3 years ago
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
3 years ago
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