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3 years ago

Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin

1 answer:

3 0

Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin

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The measures of the obtuse angles in the isosceles trapezoid are five more than four times the measures of the acute angles. Wri

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Answer:

a. i. x + y = 180 (1) and x - 4y = 5 (2)

ii. The two acute angles are 35° each and the two obtuse angles are 145° each.

Step-by-step explanation:

a. The measures of the obtuse angles in the isosceles trapezoid are five more than four times the measures of the acute angles. Write and solve a system of equations to find the measures of all the angles.

i. Write a system of equations to find the measures of all the angles.

Let x be the obtuse angles and y be the acute angles.

Since we have two obtuse angles at the top of the isosceles trapezoid and two acute angles at the bottom of the isosceles trapezoid, and also, since the sum of angles in a quadrilateral is 360, we have

2x + 2y = 360

x + y = 180 (1)

Its is also given that the measures of the obtuse angles in the isosceles trapezoid are five more than four times the measures of the acute angles.

So, x = 4y + 5 (2)

x - 4y = 5 (2)

So, our system of equations are

x + y = 180 (1) and x - 4y = 5 (2)

ii. Solve a system of equations to find the measures of all the angles.

Since

x + y = 180 (1) and x - 4y = 5 (2)

Subtracting (2) from (1), we have

x + y = 180 (1)

x - 4y = 5 (2)

5y = 175

dividing both sides by 5, we have

y = 175/5

y = 35°

From (1), x = 180° - y = 180° - 35° = 145°

So, the two acute angles are 35° each and the two obtuse angles are 145° each.

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