Line RS bisects PQ at point R, Find RQ if PQ = 14 cm
1 answer:
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<u>Answer-</u>
<em>The interval where the function is diffrentiable is </em>
<u>Solution-</u>
The given expression is,
The function will be differentiable where it is continuous and it will not be differentiable, where the function is not continuous.
The function continuous everywhere except at x = 1, because
at x = 1, its limit does not exist.
Therefore, apart from x=1, this function is differentiable everywhere. The interval will be
Answer:
65 square units
Step-by-step explanation:
Assuming point O refers to the origin, the segment OA has slope ...
m = Δy/Δx = 5/1 = 5
Then the slope of the perpendicular line AP will be the negative reciprocal of this, -1/5.
The x-intercept of the line through (1, 5) with slope -1/5 can be found by setting y=0 in the point-slope equation for that line:
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
0 -5 = (-1/5)(x -1)
25 = x -1 . . . . . . . multiply by -5
x = 26 . . . . . . . add 1
This means ΔOAP has a base length (OP) of 26 units and an altitude of 5 units. Its area is given by the formula ...
A = 1/2bh
A = 1/2(26)(5) = 65 . . . . square units
Triangle OAP has an area of 65 square units.
