The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
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You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
For the answer to the question above, the easiest way to determine is changing every runner's speed into the same unit.
<span>First = 10 m/s </span>
<span>Second = 10 miles/min = 16090.34 / 60 m/s (As 1 mile = 1609.34 meter and 1 min = 60 sec) </span>
<span>Second = 260.82 m/s </span>
<span>Third = 10 cm/hr = 10*(0.01)/60*60 (As 1 cm = 0.01 m and 1 hr = 60*60 sec) </span>
<span>Third = 0.000028 m/s </span>
<span>Fourth = 10 km/sec = 10*1000 m/s (As 1 km = 1000 m and time is already in sec) </span>
<span>Fourth = 10000 m/s </span>
<span>So fastest would be the one who covers the largest distance in 1 sec. It would be the fourth one.</span>
Answer:
The answer is option 2
Step-by-step explanation:
Working out above
Answer:
let cp be x
sp= 115x/100 ---(when 15%gain)
sp = 88x/100------(when 12% loss
according to question
115x/100 - 88x/100 = 81
or, (115x-88x)= 8100
or, 27x = 8100
or, x= 8100/27
x = 300
hence cp is RS 300
You would do
(1.1)(80,000,000)=88,000,000
That is the total cars for year two. To find the increase you would do
(1.1)(88,000,000)=96,800,000
Then you would subtract year 2 from year three
96,800,000-88,000,00=8,800,000
That gives you ur answer.
