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erica [24]
3 years ago
12

These triangles are NOT similar! Why not? Use calculations AND words to explain.

Mathematics
1 answer:
Charra [1.4K]3 years ago
4 0

Answer:

Similar triangles are (roughly speaking) triangles that have the same angles and the same ratio of the length of their sides

We can use ratios to show that the triangles are not similar

14:20 as a fraction is 14/20 as a decimal is 0.7

12:16 as a fraction is 12/16 as a decimal is 0.75

The side lengths are not the same proportion as shown by the different decimal, so the triangles are not similar

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4 0
2 years ago
15. What is 3% of 4.14 x 10?
Slav-nsk [51]

Answer:

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5 0
1 year ago
Read 2 more answers
A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successi
My name is Ann [436]

Answer:

option C

c. least: $101

greatest: $1001

Step-by-step explanation:

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successive prize.

Sum of 10 prizes = 5510

100 is the difference. there is a common difference d=100

So its a arithmetic sequence

the sum formula for arithmetic sequence is

S_n = \frac{n}{2}(2a_1 +(n-1)d)

Sn = 5510, n=10 n d= 100 we need to find out first term a1

5510 = \frac{10}{2}(2a_1 +(10-1)100)

5510 = 5 (2a1 + 900)

5510 = 10a1 + 4500

Subtract 4500 on both sides

1010= 10a1

divide by 10 on both sides

a1 = 101

so first term that is least term is 101

To find out greatest term we use formula

a_n = a_1 + (n-1) d

a(10) = 101 + (10-1)100

= 101 + 900= 1001

greatest is 1001

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
2 years ago
Apollo traveled 84 leagues in 6 days. Then he doubled his rate. How long would it take him
LenKa [72]
168 leagues in 6 days
4 0
3 years ago
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