Determine whether the following pair of lines is parallel, perpendicular or neither 2x+3y=3: 2y=3x-4

1 answer:

8 0

First, you have to simplfy your equations, 2x+3y=3. Subtract 2x from both sides, get 3y=-2x+3. Divide each side by 3, get y=-2/3x+1.
Simplify your other equation, 2y=3x-4. Divide each side by 2, get y=3/2x-2.
Now you have your 2 equations, y=-2/3x+1, and y=3/2x-2. These lines are perpendicular if you make a graph and plot the lines.

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Is the relationship between the variables in the table a direct variation, an inverse variation, or neither? If it is a direct o

inn [45]

Answer: neither direct variation nor inverse variation.

Explanation:

1) The relation between two variables, y and x, is a direct variation if and only if the quotient between them, y/x, is constant. This is: y / x = k or, equivalently, y = kx.

Note that if y / x is constant, x / y is also constant.

In a direct variation, when x increases, y increases, and when x decreases, y decreases.

2) The relation between two variables, y and x, is an inverse variation if and only if their product, y×x is constant. This is: y × x = k or, equivalently y = k /x or x = k / y.

In an inverse variation when one of the variables increases the other decreases.

3) Writhe the given table and study whether the conditions for direct or inverse variation are met:

x y y/x y×x

-6 -72 -72/(-6) = 12 (-6)(-72) = 432

-4 -47 -47 / (-4) = 11.75 (-4)(-47) = 188

-3 -36 -36 / (-3) = 12 (-3)(-36) = 108

1 12 12 / 1 = 12 (1)(12) = 12

Conclusions;

a) Since neither y / x nor y×x have the same result for every pair, the relation is neither direct nor inverse.

b) Note that if the third pair were (-4, -48) instead of (-4, -47), y / x would be 12, which make a direct variation. In this case the function that modeled it would be:

y = 12x.

8 0

4 years ago

A person 6ft tall stands 10ft from point P directly beneath a lantern hanging 30 ft above the ground. The lantern start to fall,

xxMikexx [17]

Answer:

$\frac{dL}{dt}=30 \frac{ft}{s}$

Step-by-step explanation:

In order to solve this it is always a good idea to start by drawing a diagram of the situation (See attached picture).

From the diagram we can see that we are dealing with similar triangles. We can use similar triangles to build an equation that relates the length of the shadow with the height of the lamp, so we get:

$\frac{L}{6}=\frac{10+L}{h}$

the height of the lamp can be found by subtracting the $16t^{2}$ distance the lamp falls in a given time t from the original 30ft the lamp was located at.