K = LMN
We divide both sides by LN:-
M = K / LN
Weight of an object=J
x=J/(379.2/150)
x=J/2.528
y=12.64/2.528
y=5
Hope this helps :)
Answer: (-1,1,-3) + 7 x t x (3,-2,-8)/
Step-by-step explanation:
The parametric vector equation for the position of particle is given as:
We have P= (-1,1-3)
Q= (-4,4,5)
So PQ vector= (-1+4,1-3, -3-5)
PQ = (3,-2,-8)
Magnitude of this vector= 
Magnitude=
So unit vector= (3, -2, -8)/
Now the equation is as:
P + 7 x t x unit vector = (-1,1,-3) + 7 x t x (3,-2,-8)/
Answer:
B. 5.25 square units
Step-by-step explanation:
Use side TV as the base and The dashed vertical segment as the height.
base = b = 7
height = h = 1.5
area = bh/2
area = 7 * 1.5/2
area = 10.5/2
area = 5.25 square units
Answer: B. 5.25 square units
I have a solution here that has a slight change in given where instead of <span>(4, 32), it is (3, 18). However, since the solution has provided explanations on each process, step-by-step, I believe that by thoroughly analyzing it, you might just answer this problem on your own!
</span>
f(x) = 2x² ← this is the parabola
f(3) = 2 * 9 = 18 → the parabola passes through A (3 ; 18), so its tangent line too
f'(x) = 4x ← this is the derivative
…and the derivative is the slope of the tangent line to the curve at x
f'(3) = 4 * 3 = 12 ← this is the slope of the tangent line to the curve at x = 3
Equation of the tangent line
The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept
You know that the slope of the tangent line is 12.
The equation of the tangent line becomes: y = 12x + b
The tangent line passes through A (3 ; 18), so these coordinates must verify the equation of the tangent line.
y = 12x + b
b = y - 12x → you substitute x and y by the coordinates of the point A (3 ; 18)
b = 18 - 36 = - 18
→ The equation of the tangent line is: y = 12x - 18
Intersection between the tangent line to the curve and the x-axis: → when y = 0
y = 12x - 18 → when y = 0
12x - 18 = 0
12x = 18
x = 3/2
→ Point B (3/2 ; 0)
Intersection between the vertical line passes through the point A and the x-axis: → when x = 3
→ Point C (3 ; 0)
The equation of the vertical line is: x = 3
Area of the region bounded by the parabola y = 2x², the tangent line to this parabola at (3 ; 18), and the x-axis.
= (area of the region bounded by the parabola y = 2x² and the x-axis) - (area of the triangle ABC)
= [∫ (from 0 to 3) of the parabola] - [(xC - xB).(yA - yC)/2]
= [∫ (from 0 to 3) 2x².dx] - [(xC - xB).(yA - yC)/2]
= { [(2/3).x³] from 0 to 3 } - { [3 - (3/2)].(18 - 0)/2 }
= [(2/3) * 3³] - { [(6/2) - (3/2)] * 9 }
= [(2/3) * 27] - { [(3/2) * 9 }
= 18 - (27/2)
= (36/2) - (27/2)
= 9/2 square unit
