Nearest 10 is according to the last digit.
if the digit is >5, it +1 to the tenth digit. If it's <5, it will be reduced to 0.
for your case, 392, 2 < 5, therefore rounded up to nearest tenth is 390.
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
=
, 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°
Answer:
- multiplying
- similar figures
- scale factor
- (x, y) (–3x, –3y) and (x, y) (0.23x, 0.23y)
- (x, y) --> (2x, 4y)
- Need diagram
- A. The line segment has become longer with endpoints D' (-12, -10) and E' (6 -8).
- Need image
- Reduction
- Enlargement
<em>good luck, i hope this helps :)</em>
Answer:
3rd choice
Step-by-step explanation:
(7y^6)(2y^-4)^2
= (7y^6)(4y^-8)
Calculate:
(7y^6) * (4y^-8)
28y^-2
Express with a positive exponent:
28 * 1/y^2