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.
The equation is:
250 - (120 ÷ x)
I hope this helps! :D
~PutarPotato
Answer:
x-intercept: (3,0)
y-intercept: (0,-4)
Step-by-step explanation:
To find the x and y-intercepts, we first need to understand what they are. X and y-intercepts are points on the line that passes through the x-axis and y-axis. When a point is an x-intercept, it passes through the x-axis. This means the x-coordinate is an integer, while the y-coordinate is always 0. This can be denoted by (x,0). When a point is a y-intercept, it passes through the y-axis. This means the y-coordinate is an integer, while the x-coordinate is always 0. This can be denoted by (0,y).
Now that we know what x and y-intercepts are, we can plug in x=0 and y=0 to find the intercepts.
x-intercept
4x-3y=12 [plug in y=0]
4x-3(0)=12 [multiply]
4x-0=12 [add both sides by 0]
4x=12 [divide both sides by 4]
x=3
---------------------------------------------------------------------------------------------------------
y-intercept
4x-3y=12 [plug in x=0]
4(0)-3y=12 [multiply]
0-3y=12 [subtract both sides by 0]
-3y=12 [divide both sides by -3]
y=-4
Therefore, the x-intercept is (3,0) and y-intercept is (0,-4).
Answer:
0.416 au
Step-by-step explanation:
Let y1=8sin(x) and y2=8cos(x), we must find the area between y1 and y2
Answer:
x = 27
y = 29
Step-by-step explanation:
