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seropon [69]
1 year ago
12

Sketch the graph: y=1/3x+2

Mathematics
1 answer:
Cerrena [4.2K]1 year ago
6 0

The slope of the line is 1/3and the y-intercept is at the (0, 2) as shown in the graph below.

<h3>Graph of a linear function</h3>

A linear function is a function that has a leading degree of 1. The standard equation for a linear function is expressed as:

y = mx + b

where

m is the slope which is equal to the rate of change of y coordinate to x-coordinates.

b is the y-intercept

Given the following equation

y= 1/3x + 2

The slope of the line is 1/3 and the y-intercept is at the(0,2) as shown in the graph below.

Learn more on equation of a line here: brainly.com/question/13763238

#SPJ1

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Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)
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Answer:

Area = 13.15 square units

Step-by-step explanation:

Let the given vertices be represented as follows:

A(2, -1, 1) = 2i - j + k

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C(0, 1, 1) = 0i + j + k

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(i) Let's calculate the vectors of all the sides:

\\AB = B - A =  (5i + j + 4k) - (2i - j + k)

AB = 5i + j + 4k - 2i + j - k                 [Collect like terms]

AB = 3i + 2j + 3k

BC = C - B =  (0i + j + k) - (5i + j + 4k)

BC = 0i + j + k - 5i - j - 4k                 [Collect like terms]

BC = -5i + 0j - 3k

CD = D - C =  (3i + 3j + 4k) - (0i + j + k)

CD = 3i + 3j + 4k - 0i - j - k                [Collect like terms]

CD = 3i + 2j + 3k

DA = A - D =  (2i - j + k) - (3i + 3j + 4k)

DA = 2i - j + k - 3i - 3j - 4k                [Collect like terms]

DA = -i - 4j - 3k

AC = C - A =  (0i + j + k) - (2i - j + k)

AC = 0i + j + k - 2i + j - k                [Collect like terms]

AC = -2i + 2j

BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)

BD = 3i + 3j + 4k - 5i - j - 4k                [Collect like terms]

BD = -2i + 2j

(ii) From the results in (i) above, we can deduce that;

AB = CD This implies that AB || CD  [AB is parallel to CD]

AC = BD This implies that AC || BD  [AC is parallel to BD]

(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram

<u>Now let's calculate the area</u>

To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.

In this case, we'll choose AB and AC

Area = |AB X AC|

Where;

AB X AC = \left[\begin{array}{ccc}i&j&k\\3&2&3\\-2&2&0\end{array}\right]

<u></u>

AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)

AB X AC = - 6i - 6j + 10k

|AB X AC| = \sqrt{(-6)^2 + (-6)^2 + (10)^2}

|AB X AC| = \sqrt{172}

|AB X AC| = 13.15

Area = 13.15 square units.

<u></u>

<u></u>

<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.

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