Transformations,, I have no clue what to do :(

ANSWE,LET two numbers be A and B then

A+B=52

A-B=14....linear equation in 2 variable

adding 2 eqns

2A=66... dividing both side by 2

A=33

and put A=33 in eqn A+B=52

B=52-33

B=19.

SO. LARGER NUMBER=33

Smaller number=19

Answer:

Step-by-step explanation:

each zero of the function will have a factor of (x - x₀)

h(x) = a(x + 3)(x + 2)(x - 1)

h(x) = a(x + 3)(x² + x - 2)

h(x) = a(x³ + 4x² + x - 6)

or the third option works if a = 1

however this equation gives us the points (0, -6) and (-1. -4), so "a" must be -2

h(x) = -2x³ - 8x² - 2x + 12

to fit ALL of the given points as it fits the three zeros and also h(0) and h(-1) so I guess that is why the given group is a partial set of solution sets

Step-by-step explanation:

We know that this particularly line can be named as :

• JK ←→

• LM ←→

This give us the following information :

• The line passes through the points J and K

• The line passes through the points L and M

One statement we can make is :

The points J , K , L and M are aligned. So the line passes through the points J , K , L and M.

We also know that given a line there are infinite planes that contain the line.

Given that the points J , K , L and M belong to the same line, we can state that :

There are infinite planes that contain the points J , K , L and M.

Answer:

Step-by-step explanation:

- A plane is oriented in a Cartesian coordinate system such that it makes an angle of ( π / 3 ) with the positive x - axis.

- A force ( F ) is directed along the y-axis as a vector < 0 , - 4 >

- We are to determine the the components of force ( F ) parallel and normal to the defined plane.

- We will denote two unit vectors: ( ) parallel to plane and ( ) orthogonal to the defined plane. We will define the two unit vectors in ( x - y ) plane as follows:

- The unit vector ( ) parallel to the defined plane makes an angle of ( 30° ) with the positive y-axis and an angle of ( π / 3 = 60° ) with the x-axis. We will find the projection of the vector onto the x and y axes as follows:

                          = < cos ( 60° ) , cos ( 30° ) >

- Similarly, the unit vector ( ) orthogonal to plane makes an angle of ( π / 3 ) with the positive x - axis and angle of ( π / 6 ) with the y-axis in negative direction. We will find the projection of the vector onto the x and y axes as follows:

- To find the projection of force ( F ) along and normal to the plane we will apply the dot product formulation:

- The Force vector parallel to the plane ( ) would be:

- Similarly, to find the projection of force ( ) normal to the plane we again employ the dot product formulation with normal unit vector (    ) as follows:

- To prove that the projected forces ( ) and ( ) are correct we will apply the vector summation of the two orthogonal vector which must equal to the original vector < 0 , - 4 >

                         .. proven