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Brums [2.3K]
3 years ago
7

What is the solution to this system

Mathematics
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

The answer is (2,3)

Step-by-step explanation:

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Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R
Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that \overline{RP} = \overline{SP}, the right triangle ΔRPS is an isosceles right triangle

∴ ∠PRS = ∠PSR = 45°

The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

∠PRQ = arctan(35/15) ≈ 66.8°

Therefore the bearing  from R to Q = 270 + 90 - 66.8 ≈ 293.2°

6 0
3 years ago
Find the solution to the system of equations. Select all that apply
PolarNik [594]

Answer:

A. (1,0)

Step-by-step explanation:

When you're solving a system of equations algebraically, you're finding the point that exists on both functions. So, when you're finding the solution graphically, you just need to find the point where the two functions intersect. In this case, it's A. (1,0)

5 0
3 years ago
Please help me I'm stuck on this one
AysviL [449]
Ijuliuxfd kjdfuk,kdfjhv lskdusd,f lsddv,dsjzh vas,dj
8 0
3 years ago
In circle L, points E and F lie on the circle such that E, F and L are not collinear. If LE, LF and EF are drawn then which of t
frozen [14]

Answer:

<u>option (2) it is and isosceles triangle. </u>

Step-by-step explanation:

In circle L, points E and F lie on the circle such that E, F and L are not collinear

If LE, LF and EF are drawn

So, LE = EF = the radius of the circle L

So, LEF is an isosceles triangle.

The answer is option (2) it is and isosceles triangle.

8 0
3 years ago
CAN SOMEONE PLEASE HELP ME!!!!!!!!!!!!!
antiseptic1488 [7]

Answer:

B

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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