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Nonamiya [84]
3 years ago
10

which table represents viable solutions for y=5x, where x is the number of tickets sold for the school play a y is the amount of

money collected for the tickets?

Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0

Given:

The given equation is:

y=5x

Where x is the number of tickets sold for the school play a y is the amount of money collected for the tickets.

To find:

The correct table of values from the given options.

Solution:

We know that the number of tickets and amount of money collected for the tickets cannot be negative. So, options A and B are incorrect.

We have,

y=5x

For x=0,

y=5(0)

y=0

For x=10,

y=5(10)

y=50

For x=51,

y=5(51)

y=255

For x=400,

y=5(400)

y=2000

In option C, all the ordered pairs (0,0), (10,50), (51,255), (400,2000) are in the table. So, option C is correct.

For x=65

y=5(65)

y=325

Since the ordered pair (65,350) does not satisfy the given equation, therefore the option D is incorrect.

Hence, the correct option is C.

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If can babysits 2 hours and she earns $14 then she works 4 hours and gets $28 what how much does she make if she does 6 hours
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Answer:

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Subtract 1.05 from a certain number. Multiply the difference by 0.8, add 2.84 to the product then divide the sum by 0.01 and get
Dima020 [189]
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6 0
3 years ago
Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy
Margaret [11]

Answer:

Part 1) The exact solutions are

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})   and  (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

y=x^{2} +3x ----> equation A

y=2x+5 ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

<em>Find the exact solutions of the system</em>

equate equation A and equation B

x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2} +x-5=0  

so

a=1\\b=1\\c=-5

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}

x=\frac{-1\pm\sqrt{21}} {2}

so

The solutions are

x_1=\frac{-1+\sqrt{21}} {2}

x_2=\frac{-1-\sqrt{21}} {2}

<em>Find the values of y</em>

<em>First solution</em>

For x_1=\frac{-1+\sqrt{21}} {2}

y=2(\frac{-1+\sqrt{21}} {2})+5

y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}

The first solution is the point (\frac{-1+\sqrt{21}} {2},4+\sqrt{21})

<em>Second solution</em>

For x_2=\frac{-1-\sqrt{21}} {2}

y=2(\frac{-1-\sqrt{21}} {2})+5

y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}

The second solution is the point (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Round to the nearest hundredth

<em>First solution </em>

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21}) -----> (1.79,8.58)

(\frac{-1-\sqrt{21}} {2},4-\sqrt{21}) -----> (-2.79,-0.58)

see the attached figure to better understand the problem

6 0
3 years ago
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