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NeX [460]
2 years ago
15

The cost of 4 cookies, 6 doughnuts, and 3 boxes of doughnut holes is $8.15. The cost of 2 cookies, 3 doughnuts, and 4 boxes of d

oughnuts holes is $7.20. What is the cost of a box of doughnut holes?
Mathematics
1 answer:
Naddika [18.5K]2 years ago
5 0

Answer:

$1.25

Step-by-step explanation:

This can best be determined using a set of linear equations that are solved simultaneously.

This pair of linear equations may be solved simultaneously by using the elimination method. This will involve ensuring that the coefficient of one of the unknown variables is the same in both equations.

Let the cost of a cookie be c, cost of a doughnut be d and that of a box of doughnut hole be h then if cost of 4 cookies, 6 doughnuts, and 3 boxes of doughnut holes is $8.15, we have

4a + 6d + 3h = 8.15

and the cost of 2 cookies, 3 doughnuts, and 4 boxes of doughnuts holes is $7.20 then

2a + 3d + 4h = 7.20

Dividing the first by 2

2a + 3d + 1.5h = 4.075

subtracting from the second equation

2.5h = 3.125

h = 1.25

The cost of a box of doughnut holes is $1.25

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Answer:

3 < c < 9

Step-by-step explanation:

The length of a side of a triangle cannot be negative. This eliminates the first and last options.

The addition of two sides of a triangle must be greater than the third side. In this case:

3 + 6 = 9 > c

So, the second option is correct. The third option is not correct, because, for example, c = 8 is possible

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Step-by-step explanation:

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2 years ago
Find a third-degree polynomial equation with rational coefficients that has roots –4 and 2 + i
ella [17]
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8 0
3 years ago
Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.
jeka94

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

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This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

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2 years ago
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