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Vesna [10]
3 years ago
5

Tracy sells water at a fair in 17oz bottles or 11oz cans. If on one shelf she has 11 containers of water that hold a total of 15

7oz of water, how many bottles and cans of water are on the shelf?
Mathematics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

Number of bottles = 6

Number of cans = c

Step-by-step explanation:

Given that:

Container for selling water :

17 oz bottle

11 oz can

Let number of bottles = b

Number of cans = c

b + c = 11 - - - (1)

17b + 11c = 157 - - - (2)

b = 11 - c

Substitute into (2)

17(11 - c) + 11c = 157

187 - 17c + 11c = 157

187 - 6c = 157

-6c = 157 - 187

-6c = -30

c = 30/6

c = 5

From ; b = 11 - c

b = 11 - 5

b = 6

Hence,

Number of bottles = 6

Number of cans = c

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Three consecutive even numbers have a sum between 84 and 96.
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Answer:

a. 84 < n + (n + 2) + (n + 4) < 96

b. 26 < n < 30


Step-by-step explanation:

Let n = 1st number

n+2   2nd  even number

n+4 = 3rd even number

The sum of these 3 numbers is

n+n+2+n+4

It must be between 84 and 96  (it does not include this 84 and 96)

84< n+n+2+n+4 < 96


Now we need to solve this

Combine like terms

84 < 3n +6< 96

Subtract 6 from all sides

84 -6 < 3n+6-6 < 96 -6

78 < 3n < 90

Divide all sides by 3

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3 years ago
Match the parabolas represented by the equations with their foci.
Elenna [48]

Function 1 f(x)=- x^{2} +4x+8


First step: Finding when f(x) is minimum/maximum
The function has a negative value x^{2} hence the f(x) has a maximum value which happens when x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2. The foci of this parabola lies on x=2.

Second step: Find the value of y-coordinate by substituting x=2 into f(x) which give y=- (2)^{2} +4(2)+8=12

Third step: Find the distance of the foci from the y-coordinate
y=- x^{2} +4x+8 - Multiply all term by -1 to get a positive x^{2}
-y= x^{2} -4x-8 - then manipulate the constant of y to get a multiply of 4
4(- \frac{1}{4})y= x^{2} -4x-8
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is 12- \frac{1}{4}=11.75

Hence the coordinate of the foci is (2, 11.75)

Function 2: f(x)= 2x^{2}+16x+18

The function has a positive x^{2} so it has a minimum

First step - x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4
Second step - y=2(-4)^{2}+16(-4)+18=-14
Third step - Manipulating f(x) to leave x^{2} with constant of 1
y=2 x^{2} +16x+18 - Divide all terms by 2
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4( \frac{1}{8}y= x^{2} +8x+9

So the distance of focus from y-coordinate is \frac{1}{8} to the north of y=-14
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: f(x)=-2 x^{2} +5x+14

First step: the function's maximum value happens when x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25
Second step: y=-2(1.25)^{2}+5(1.25)+14=17.125
Third step: Manipulating f(x)
y=-2 x^{2} +5x+14 - Divide all terms by -2
-2y= x^{2} -2.5x-7 - Manipulate coefficient of y to get a multiply of 4
4(- \frac{1}{8})y= x^{2} -2.5x-7
So the distance of the foci from the y-coordinate is -\frac{1}{8} south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when x=8.5 and y=79.25. The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when x=-2.75 and y=-10.125. Manipulating coefficient of y, the distance of foci from y-coordinate is \frac{1}{8} to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when x=1.5 and y=9.5. The distance of the foci from the y-coordinate is \frac{1}{8} to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

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