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VLD [36.1K]
3 years ago
7

15 more than 3 times a number is at most 7

Mathematics
1 answer:
hammer [34]3 years ago
8 0

Answer:

Let x = the number

"Fifteen more than" means addition

"Three times a number" is multiplication

"three less than" is subtraction, as is the term difference

Putting it all together, we have

7 - (15 + 3x) = (9 - 5x) - 3

-8 + 3x = 6 - 5x

-8 = 6 - 8x

-14 = -8x

14/8 = 7/4 = x

Step-by-step explanation:

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Suppose three riders rode a total of 182 miles. If they used a total of 13 horses, and rode each horse the same number of miles,
umka21 [38]

Answer:

They rode 14 miles before replacing each horse

Step-by-step explanation:

We will be working under the assumption that all three riders sat on one horse at a time and rode it while the other horses rested.

From the problem, we can understand that the horses were each ridden for the same distance. This means that to get the total distance a horse rode before it was changed, we can divide the total distance by the number of horses that were used for the journey.

Distance each horse rode = 182/ 13 = 14 miles.

Therefore, each horse was ridden for 14 miles before it was changed.

5 0
3 years ago
Simon is creating a flower arrangement. A client asked him to select 4 of the 7 available flower options to create the arrangeme
SashulF [63]

Answer:

i dont know but good luck! <3

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Write 9,167.364 in expanded form
beks73 [17]

Answer:

4 + 60 + 300 + NaN + 70000 + 600000 + 1000000 + NaN + 900000000

7 0
2 years ago
9(d − 93) = –36 d = _______
goldenfox [79]

Steps to solve:

9(d - 93) = -36d

~Distribute

9d - 837 = -36d

~Subtract 9d to both sides

-837 = -45d

~Divide -45 to both sides

18.6 = d

Best of Luck!

4 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
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