Generate Quiz 6A332E4094FD5 - PLS Academy

You have 30 minutes to complete the quiz

Generate Quiz 6A332E4094FD5

1 / 10

If ABC is isosceles triangle, then:

m1=m2

m2=m3

m1≠m2≠m3

In a triangle, no slope is equal to any other.

2 / 10

The slope of a line that makes an angle of measure 30 with y-axis is

-3

√3

±1/√3

±√3

3 / 10

[x²+1/x²-1] wr.to x²+1/x²-1:

x2+1/x2-1

2[x2+1/x2-1]

x2-1/x2+1

2[x2+1/x2-1]

4 / 10

The minimum value of the function ∅(x,y) = 4x + 6y under the constraints 2x – 3y ≤ 6, 2x + y ≥ 2, 2x + 3y ≤ 12,

x ≥ 0, y ≥ 0 exists at the corner point:

(0,4)

(9/2,1)

Both A and B

None of these

The limitations that are specified are

2x – 3y ≤ 6

2x + y ≥ 2

2x + 3y ≤ 12.

x ≥ 0.

y ≥ 0.

These constraints establish the following corner points of the feasible region:

A: (0,4)

B: (9/2,1)

C: 2x – 3y = 6 and 2x + y = 2 intersect (solving these two equations yields (2,0), but since y is negative, this point is not possible).

D: The intersection of 2x + y = 2 and 2x + y = 12 results in (6/5,16/5), but since x is negative, this point is not possible.

E: 2x + 3y = 12 and 2x – 3y = 6 intersect; solving these two equations yields (9/4,3/4)

The objective function ∅(x,y) = 4x + 6y should now be assessed at each of these corner points:

A: ∅(0,4) = 4(0) + 6(4) = 24

B: ∅(9/2, 1) = 4(9/2) + 6(1) = 18 + 6 = 24

C: Not possible

D: Not possible

E: ∅ (9/4, 3/4) = 4(9/4) + 6(3/4) = 9 + 9/2 = 27/2

Consequently, the corner points (0,4) and (9/2,1) produce a minimum value of 24 for the objective function, and the minimum value of the function ∅(x,y) = 4x + 6y under the specified constraints occurs at these points.

The limitations that are specified are

2x – 3y ≤ 6

2x + y ≥ 2

2x + 3y ≤ 12.

x ≥ 0.

y ≥ 0.

These constraints establish the following corner points of the feasible region:

A: (0,4)

B: (9/2,1)

C: 2x – 3y = 6 and 2x + y = 2 intersect (solving these two equations yields (2,0), but since y is negative, this point is not possible).

D: The intersection of 2x + y = 2 and 2x + y = 12 results in (6/5,16/5), but since x is negative, this point is not possible.

E: 2x + 3y = 12 and 2x – 3y = 6 intersect; solving these two equations yields (9/4,3/4)

The objective function ∅(x,y) = 4x + 6y should now be assessed at each of these corner points:

A: ∅(0,4) = 4(0) + 6(4) = 24

B: ∅(9/2, 1) = 4(9/2) + 6(1) = 18 + 6 = 24

C: Not possible

D: Not possible

E: ∅ (9/4, 3/4) = 4(9/4) + 6(3/4) = 9 + 9/2 = 27/2

The limitations that are specified are

2x – 3y ≤ 6

2x + y ≥ 2

2x + 3y ≤ 12.

x ≥ 0.

y ≥ 0.

These constraints establish the following corner points of the feasible region:

A: (0,4)

B: (9/2,1)

C: 2x – 3y = 6 and 2x + y = 2 intersect (solving these two equations yields (2,0), but since y is negative, this point is not possible).

D: The intersection of 2x + y = 2 and 2x + y = 12 results in (6/5,16/5), but since x is negative, this point is not possible.

E: 2x + 3y = 12 and 2x – 3y = 6 intersect; solving these two equations yields (9/4,3/4)

The objective function ∅(x,y) = 4x + 6y should now be assessed at each of these corner points:

A: ∅(0,4) = 4(0) + 6(4) = 24

B: ∅(9/2, 1) = 4(9/2) + 6(1) = 18 + 6 = 24

C: Not possible

D: Not possible

E: ∅ (9/4, 3/4) = 4(9/4) + 6(3/4) = 9 + 9/2 = 27/2

5 / 10

The centroid of the triangle whose vertices are (4,-2) (-2,4) (5,5) is:

(11/3,11/3)

(11/3,7/3)

(7/3,7/3)

(7,7)

6 / 10

7 / 10

if x=asinθ, y=bcos θ then d2y/dx2 is equal to

-b/a sec2θ

b/a2 sec2 θ

-b/a2 sec3θ

a/b2sec2θ

8 / 10

unit vector of v is defined as

v/|v|

|v|/v

v.|v|

A vector with the same direction as v but a magnitude of 1 is called the unit vector of a vector v. v/|v| is the formula for the unit vector, where |v| is the vector v’s magnitude (length).

9 / 10

The angle between the lines √3x-y-2=0 and x-3y+1=0 is

15'

10 / 10

The line y=mx+c touches the parabola x² = 4ay if

c= -a /m2

c=-am²

c=- a/m

c=-am

answer:

The right response is b.

This question can be answered by taking into account the requirement that a line and a parabola meet at a single place.

A tangent is the point at where a line and a parabola intersect. Stated differently, the line equation and the parabola equation can be equated to create a quadratic equation with equal roots.

How to solve it:

In the parabola equation, replace y with the line equation.

We will then have a quadratic equation in x as a result.

The discriminant (b^2-4ac) must equal zero in order for the roots of the quadratic equation to be equal.

Determine c by solving the discriminant equation.

Finding c by solving the equation 16a2m2+16ac=0

We obtain: c=−am2.

answer:

The right response is b.

This question can be answered by taking into account the requirement that a line and a parabola meet at a single place.

A tangent is the point at where a line and a parabola intersect. Stated differently, the line equation and the parabola equation can be equated to create a quadratic equation with equal roots.

How to solve it:

In the parabola equation, replace y with the line equation.

We will then have a quadratic equation in x as a result.

The discriminant (b^2-4ac) must equal zero in order for the roots of the quadratic equation to be equal.

Determine c by solving the discriminant equation.

Finding c by solving the equation 16a2m2+16ac=0

We obtain: c=−am2.

Your score is

PLS Academy App!

QUICK LINKS

PREP

COURSES

SUPPORT