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3 years ago

9

How would you graph the solution to -5w + 9 = 14 on a number line?

1 answer:

kirill [66]3 years ago

3 0

Answer:

Draw a point one unit to the left of zero.

Step-by-step explanation:

The equation of unknown w is given here and we have to solve it and then we have to check how to draw the point w on the number line.

Now, the equation is - 5w + 9 = 14

⇒ - 5w = 5

⇒ w = - 1

Now, we know that in a number line if we move right from zero then this is a positive direction and if we move left from zero then this direction is the negative direction.

Hence, to plot w = - 1, we must draw a point one unit to the left of zero. (Answer)

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3 years ago

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Answer:

2) ABCD is a RECTANGLE.

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Step-by-step explanation:

Here, the given points of the quadrilateral are:

A(6, –4), B(11, –4), C(11, 6), D(6, 6)

Now, if P(a,b) and Q(c,d) are any two given points, then the distance between them is given by DISTANCE FORMULA as:

$PQ = \sqrt{(c-a)^2 + (d-b)^2$

Use this to find the length of all 4 sides:

AB with coordinates A(6, –4), B(11, –4) is given as:

$AB = \sqrt{(11-6)^2 + (-4-(-4))^2} = \sqrt{(5)^2 + 0} = 5$

⇒ AB = 5 units

BC with coordinates B(11, –4), C(11, 6) is given as:

$BC = \sqrt{(11-11)^2 + (6-(-4))^2} = \sqrt{(0)^2 + 10^2} = 10$

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CD with coordinates C(11, 6), D(6, 6) is given as:

$CD = \sqrt{(11-6)^2 + (6-6)^2} = \sqrt{(5)^2 + 0} = 5$

⇒ CD = 5 units

AD with coordinates A(6, –4),D(6, 6) is given as:

$AD = \sqrt{(6-6)^2 + (-4-(-4))^2} = \sqrt{(0)^2 + 10^2} = 10$

⇒ AD = 10 units

Now, here AB = CD = 5 units, AD = BC = 10 units

Also, in a RECTANGLE, OPPOSITE SIDES ARE EQUAL IN LENGTH.

Hence, ABCD is a RECTANGLE.

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fredd [130]

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Answer:

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Step-by-step explanation:

The detailed steps is shown in the attachment.

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