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3 years ago

Which is the graph of the linear equation y = –one-fourthx + 3?

Mathematics

2 answers:

quester [9]3 years ago

8 0

Answer:

Step-by-step explanation:

Graphed in Desmos!

Sophie [7]3 years ago

5 0

Explanation: The slope-intercept form is y=mx+b y = m x + b , where m is the slope and b is the y-intercept. Using the slope-intercept form, the y-intercept is −3 .

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Three girls search for geodes in the desert. corinne collected 11 1/8 pounds, ellen collected 4 5/8 pounds, and leonda collected

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3 years ago

What is the value of [-4.6]

solniwko [45]

<h2>Answer:</h2>

The value is:

-4

<h2>Step-by-step explanation:</h2>

We are asked to find the value of the ceiling function: $\left \lceil -4.6 \right \rceil$

As we know that the ceiling function always occupy the higher value in integers.

i.e. the ceiling function act as follows:

it takes value 0 when -1< x≤0

1 when 0 < x ≤ 1

2 when 1 < x ≤2

and so on.

As we know that:

-4.6 lie between -5 and -4.

Hence, we have:

$\left \lceil -4.6 \right \rceil=-4$

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3 years ago

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Nana76 [90]

Hello!

Two rays make any angle

The answer is C)2

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3 years ago

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A survey asked people which kind of pet they owned. The results are shown in the table below. Of the women who were surveyed, wh

bagirrra123 [75]

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3 years ago

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The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago