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3 years ago

How do I Graph y= –73x+2 .

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

Answer:

See attachment

Step-by-step explanation:

You can obtain any two points on the graph of $y=-73x+2$ and use it to draw its graph.

When x=0, $y=-73*0+2=2$

So you plot (0,2)

When x=1, $y=-73*1+2=-71$

You again plot (1,-71).

With a straight edge you can now draw a straight line through the two points.

See attachment

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Answer:

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Step-by-step explanation:

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3 years ago

Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral

Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

$l_{AB} = \sqrt{\left (6-4 \right )^{2}+\left (6-1 \right )^{2}} = \sqrt{29}$

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

$l_{BC} = \sqrt{\left (1-6 \right )^{2}+\left (4-6 \right )^{2}} = \sqrt{29}$

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

$l_{CD} = \sqrt{\left (-1-1 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}$

The length of side DA, with D = (-1, -1) and A = (1,4) gives;

$l_{DA} = \sqrt{\left (4-(-1) \right )^{2}+\left (1-(-1) \right )^{2}} = \sqrt{29}$

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore;

$Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1$

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

$Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1$

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

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