I think it's A because of PEMDAS so you multiply 8 by n-3
If I am reading this right, it looks like the 10, 3, 2, 1 are Adjustments and the Adjusted TB should equal the difference. Make sure you know how to add and subtract the debit and credit adjustments correctly.
TB +/- Adj = ATB
<h3>
Answer:</h3>
see attached for a graph
domain and range: all real numbers
<h3>
Step-by-step explanation:</h3>
The function is written in slope-intercept form, showing that it has a slope of -3 and a y-intercept of +7. The y-intercept (0, 7) is a point on the line, as is a point 1 unit to the right and down 3 units, (1, 4).
The graph will be the line through these two points.
_____
As with any odd-degree polynomial function, both domain and range are all real numbers: (-∞, ∞).
Answer:
(a) 315°
(b) 3°
(c) 238°
Step-by-step explanation:
Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.
<h3>(a)</h3>
The bearing of P from R is 180° different from the bearing of R from P it will be ...
135° +180° = 315° . . . . bearing of P from R
__
<h3>(b)</h3>
The bearing of Q from R is 48° more than the bearing of P from R, so is ...
315° +48° = 363°, or 3° . . . . bearing of Q from R
__
<h3>(c)</h3>
The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...
180° -48° -55° = 77°
The bearing of Q from P is 77° less than the bearing of R from P, so is ...
135° -77° = 58°
As above, the reverse bearing from Q to P is ...
58° +180° = 238° . . . . bearing of P from Q