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mr Goodwill [35]
3 years ago
11

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of points C are (0, -3). The origin is the mid –point

of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

Answer

Point b is (0,0) Point a is (3,-2) Im not doing the second part

Step-by-step explanation:

You might be interested in
50
ozzi

Answer:

90 km, N 46° E

Step-by-step explanation:

<em>A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.</em>

Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).

This will be the jet's distance from its starting point.

In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.

The <u>Law of Cosines</u> will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.

In this case, the known angle is 110° (angle B) so we will use the <u>Law of Cosines</u> respective to B.

  • b² = a² + c² - 2ac cosB

Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).

  • b² = (60)² + (50)² - 2(60)(50) cos(110°)
  • b² = 6100 -(-2052.12086)
  • b² = 8152.12086
  • b = 90.28909602
  • b ≈ 90 km

The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.

The <u>Law of Cosines</u> with respect to A:

  • a² = b² + c² - 2bc cosA

Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).

  • (60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
  • 3600 = 8152.12086 - 6528.909602 cosA
  • -4552.12086 = -6528.909602 cosA
  • 0.6972252853 = cosA
  • A = cos⁻¹(0.6972252853)
  • A = 45.79519
  • A ≈ 46°

The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.

7 0
3 years ago
Sonja has 5 apples in her pantry, she wants all 8 of her horses to eat the same amount as a treat. How much will each horse eat?
Mandarinka [93]

Answer:1

Step-by-step explanation:

8÷5=1.6 but you don’t want the 6 or the decimal (.) so it’s 1!

5 0
4 years ago
Read 2 more answers
PLEASE HELP ITS DUE TMR
sattari [20]

Answer:

it's 42 my bad...

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Olivia's class collected 3,591 bottle caps in 57 days. The same number of bottle caps were collected each day. How many bottle c
irinina [24]
This is a great question for a calculator

But anyway all you have to do is divide 3,591 by 57 to find your answer which the answer is 63 bottle caps per day
3 0
3 years ago
Read 2 more answers
In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline
VLD [36.1K]

Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>

6 0
3 years ago
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