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3 years ago

Find x & y so that quadrilateral is a parallelogram. (Look at image) please help!!

Mathematics

1 answer:

dybincka [34]3 years ago

4 0

Answer:

y = 2 and x = -3

Step-by-step explanation:

we know that opposite sides of a parallelogram are equal so

7y + 3 = 12y - 7

12y - 7y = 3 + 7

5y = 10

y = 10 / 5

y = 2

Now

-6x = -4x + 6

-6x + 4x = 6

-2x = 6

x = 6 / - 2

x = - 3

Hope it will help :)❤

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skelet666 [1.2K]

Answer:

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djverab [1.8K]

Answer:

Step-by-step explanation:

To sketch a quadratic function we need two things:

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2) Vertex

3) y-intercept

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$y=a(x-h)^2+k$

where,

a represents the nature of the graph it can be maximum or minimum , meaning, if a > 0 then minimum(u shaped curve/happy face) and if a < 0 then maximum(n shaped curve/sad face).

h represents the x-coordinate of the vertex.

k represents the y-coordinate of the vertex.

Now if we compare g(x) with our completing square form we get the following:

$g(x)=(x-4)^2+12\\y=a(x-h)^2+k\\$

When we simply compare the following we get ,

a = 1 , which means a > 0 since 1 is greater than 0 the nature of the curve will be minimum(happy face/u shaped)

h = 4, which means the x-coordinate of the vertex is 4

k = 12, which means the y-coordinate of the vertex is 12

Now we have the nature of the curve, we have the vertex now all we need is the y-intercept.

For y-intercept:

For y-intercept meaning at which point will the graph cross the y-axis(0 , y)

For that we expand the formula and turn it into the standard quadratic equation form by using the formula (a - b)^2

$y=(x-4)^2+12\\y=((x)^2-2(x)(4)+(4)^2)+12\\y=x^2-8x+16+12\\y=x^2-8x+28\\$

now we compare with the standard quadratic form:

$y=ax^2+bx+c$

here c is the y-intercept and while comparing we can see that c = 28 ,

so the curve cuts the y-axis at (0 , 28)

So we have all the three things that we need to graph our function.

So we just plot the y-intercept , the vertex , and join the dots. Just a tip draw a dotted line on the x-coordinate of the vertex because the vertex point is also called as a turning point where the graph goes in the opposite direction just like a mirror reflection. I attached 2 images you can check them out. One is handmade(i know i suck at drawing but still xD) , one is sketched by online graphing calculator.

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Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

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