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Sveta_85 [38]
3 years ago
7

Figure ABCD has vertices A(−3, 2), B(2, 2), C(2, −4), and D(−3, −2). What is the area of Figure ABCD? 5 square units 20 square u

nits 25 square units 40 square units

Mathematics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

The area of Figure ABCD is 25 square units

Step-by-step explanation:

we have

A(−3, 2), B(2, 2), C(2, −4), and D(−3, −2)

step 1

Plot the figure

using a graphing tool

The figure is a trapezoid

see the attached figure

step 2

Find the area of the trapezoid

The area of the trapezoid is equal to

A=\frac{1}{2}[AD+BC]AB

we have

AD=2-(-2)=4\ units ---> difference in the y-coordinates

BC=2-(-4)=6\ units ---> difference in the y-coordinates

AB=2-(-3)=5\ units ---> difference in the x-coordinates

substitute

A=\frac{1}{2}[4+6]5

A=25\ units^{2}

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0.0618.

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14-18= -4

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Read 2 more answers
Solve x2+4x+6=0 by graphing
klemol [59]

Answer: x = -2±√2 i or no real roots. See the explanation below.

Step-by-step explanation:

If you draw the graph, you will notice that there are no common points or you can say that there are no x-intercepts.

This means that the equation doesn't have any real roots.

If there are 2 common points (2 x-intercepts) then there are 2 real roots (answers.)

If there is only one common point (vertex is on x-axis) then there is only 1 real root.

If there aren't any common points (no x-intercepts) then there are no real roots.

So first, I'm gonna convert the equation into function and convert the standard form to vertex form.

x^2+4x+6=0\\y=x^2+4x+6\\y=(x^2+4x+4)-4+6\\y=(x+2)^2+2

We notice that the vertex is at (-2,2), the graph doesn't have x-intercepts (Graph below)

So the answer is clearly complex.

If you solve this by solving the equation then it'll be.

x^2+4x+6=0\\x= \frac{-b+-\sqrt{b^2-4ac}}{2a} \\x=\frac{-4+-\sqrt{16-24} }{2}\\x=\frac{-4+-\sqrt{-8} }{2}\\\\x=\frac{-4+-2\sqrt{2}i }{2}\\\\x={-2+-1\sqrt{2}i }\\\\

The answer is complex.

However, you can find the value of x easily by graphing. For example.

x^2+6x+5=0

Factor then you get

(x+5)(x+1)=0\\x=-5,-1

Or you can convert it to vertex form then graph.

y=x^2+6x+5

y=(x^2+6x+9)-9+5\\y=(x+3)^2-4

The vertex is at (-3,-4) then graph parabola (x-intercepts determine the roots of quadratic equation. For the graph, parabola intercepts x at (-5,0) and (-1,0) that means both -5 and -1 both are the value of x.

Note:

From the standard form, y=ax^2+bx+c The graph is called Parabola, and it's Quadratic Function.

a is how the graph is, for example. If a>0, the graph is supine parabola and if a<0 then the graph is inverted parabola.

|a| is determined if the graph is wide or narrow, the more the value of a is, the more narrow it will be. The less the value of a is, the more wide it will be.

b is what determined the change of graph (likely vertex). Without b, then it'd be ax^2+k which only move y-axis and not x-axis.

c is what determined the y-intercept and y-axis. If there is no b or the value of b is 0 then the vertex is at (0,c). If there is the value of b then c is only determined the y-intercept.

4 0
3 years ago
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