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padilas [110]
3 years ago
7

What is the original price if there is a 10% discount and the sale price is 76.50?

Mathematics
2 answers:
Elden [556K]3 years ago
4 0
Let the original price = x
then  76.50 = 0.90x
x = 76.50 / 0.90 = 85 
Allisa [31]3 years ago
3 0
This being algebra, your best bet would be to represent the "original price" by a letter, such as n.

If there's a 10% discount on this item, the price you'd pay would be (1-0.10)n, and this price is $76.50

Solve (1-0.10)n = $76.50 for n.


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Vinil7 [7]

Answer:

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8 0
3 years ago
What number is 53% of 470?
Ivahew [28]
53% of 470 is exactly 249.10000000000002. What ever is closest to that is your answer.

3 0
3 years ago
How do I solve this?Help me please!!!
NISA [10]
Given:
Ship M travels E 15 km, then N35E 27 km. Its sub travels down 48° 2 km from that location.

Ship F travels S75E 20 km, then N25E 38 km. The treasure is expected to be at this location 2.18° below horizontal from the port.

Find:
1a. The distance from port to Ship M
1b. The distance from port to the sub
1c. The angle below horizontal from the port to the sub

2a. The distance from port to Ship F
2b. The depth to the expected treasure location
2c. The distance from port to the expected treasure location

Solution:
It can be helpful to draw diagrams. See the attached. The diagram for depth is not to scale.

There are several ways this problem can be worked. A calculator that handles vectors (as many graphing calculators do) can make short work of it. Here, we will use the Law of Cosines and the definitions of Tangent and Cosine.

Part 1
1a. We are given sides 15 and 27 of a triangle and the included angle of 125°. Then the distance (m) from the port to the ship is given by the Law of Cosines as
  m² = 15² +27² -2·15·27·cos(125°) ≈ 1418.60
  m ≈ 37.66
The distance from port to Ship M is 37.66 km.

1b. The distance just calculated is one side of a new triangle with other side 2 km and included angle of 132°. Then the distance from port to sub (s) is given by the Law of Cosines as
  s² = 1418.60 +2² -2·37.66·2·cos(132°) ≈ 1523.41
  s ≈ 39.03
The distance from port to the sub is 39.03 km.

1c. The Law of Sines can be used to find the angle of depression (α) from the port. That angle is opposite the side of length 2 in the triangle of 1b. The 39.03 km side is opposite the angle of 132°. So, we have the relation
  sin(α)/2 = sin(132°)/39.03
  α = arcsin(2·sin(132°)/39.03) ≈ 2.18°
The angle below horizontal from the port to the sub is 2.18°.

Part 2
2a. We are given sides 20 and 38 of a triangle and the included angle of 100°. Then the distance (f) from the port to the ship is given by the Law of Cosines as
  f² = 20² +38² -2·20·38·cos(100°) ≈ 2107.95
  f ≈ 45.91
The distance from port to Ship F is 45.91 km.

2b. The expected treasure location is at a depth that is 2.18° below the horizontal from the port. The tangent ratio for an angle is the ratio of the opposite side (depth) to the adjacent side (distance from F to port), so we have
  tan(2.18°) = depth/45.91
  depth = 45.91·tan(2.18°) ≈ 1.748
The depth to the expected treasure location is 1.748 km.

2c. The distance from port to the expected treasure location is the hypotenuse of a right triangle. The cosine ratio for an angle is the ratio of the adjacent side to the hypotenuse, so we have
  cos(2.18°) = (port to F distance)/(port to treasure distance)
  (port to treasure distance) = 45.91 km/cos(2.18°) ≈ 45.95
The distance from the port to the expected treasure is 45.95 km.

Part 3
It seems the Mach 5 Mimi is the ship most likely to have found the treasure. That one seems ripe for attack. Its crew goes to a location that is 2.18° below horizontal. The crew of the FTFF don't have any idea where they are going. (Of course, the pirate ship would have no way of knowing if it is only observing surface behavior.)

5 0
3 years ago
I need help so somebody plz help me
Anettt [7]

Answer: A

This is simple. Absolute value inequalities can NOT have a negative. For D, the absolute value is not completely simplified. The negative sign will be removed in the process of solving it.

5 0
3 years ago
A rectangular box is to have a square base and a volume of 40 ft3. If the material for the base costs $0.36/ft2, the material fo
blagie [28]

Answer:

  • length: 2 ft
  • width: 2 ft
  • height: 10 ft

Step-by-step explanation:

The total cost of the top and bottom is $0.36 + 0.14 = $0.50 per square foot.

The total cost of a pair of opposite sides is $0.05 +0.05 = $0.10 per square foot.

A minimum-cost box will have the cost of any pair of opposite sides be the same. Here, that means the box will have a side area that is 5 times the area of the top or bottom.

Since the base is square, that means the box is the shape of 5 cubes stacked one on the other. Each of those would be 8 ft³, so would have edge dimensions of ∛8 = 2 feet. The height is 5 times that, or 10 ft.

The box is 2 feet square by 10 feet high:

  • length: 2 ft
  • width: 2 ft
  • height: 10 ft

_____

If you feel the need to write an equation, you can let x represent the edge length of the base. Then the cost of the top and bottom will be ...

  top&bottom cost = 0.50·x²

The height of the box is 40/x², so the cost of the four sides will be ...

  side cost = (0.05)(4x)(40/x²) = 8/x

This is minimized when the derivative of the sum of these costs is zero:

  cost = top&bottom cost + side cost

  cost = 0.50x² + 8/x

  d(cost)/dx = 1.00x -8/x² = 0

Multiplying by x², we get ...

  x³ -8 = 0

  x = ∛8 = 2 . . . . . . as above

  height = 40/x² = 40/4 = 10 . . . . . as above

6 0
3 years ago
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