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dangina [55]
3 years ago
9

Someone please help me with this!!

Mathematics
1 answer:
babunello [35]3 years ago
5 0

A= (3.5,3.5)

b=(3.5,7)

c=(10.5,7)

D=(10.5,3.5)

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-3x - 3y - z= -6
kifflom [539]

Answer:

Ammm, I think the answer you have is wrong, but what I really think is that the equations you put in are not the right ones because the answer for these is "ugly" (decimals etc...). Want to check your equations?

Step-by-step explanation:

Here is a step by step explanation to your equations (I assumed +3z for the second but you can change it). You can edit it for your equations if you want to.

https://simplisico.com/share/q/QiWKOyN5uYgkSG4iBXOmdG0i

5 0
2 years ago
Please subscribe to my friends channel <br> channel is called morrison pereira
neonofarm [45]

Answer:

ok

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
This si more tiring than actually doing the work
DENIUS [597]

Answer:

D(WZ║XY)

Step-by-step explanation:

The definition of a trapezoid is "a quadrilateral with only one pair of parallel sides".  WX and ZY are obviously not parrelel, so you need WZ and XY to be parellel to follow the criteria of a trapezoid.

6 0
3 years ago
Point C is on line segment \overline{BD} BD . Given CD=x,CD=x, BC=5x-5,BC=5x−5, and BD=2x+7,BD=2x+7, determine the numerical len
NARA [144]

Answer:

CD = 3

Step-by-step explanation:

Given

CD = x

BC = 5x - 5

BD = 2x + 7

Required

Determine CD

Since, C is a point on BD, the relationship between the given parameters is;

BD = BC + CD

Substitute the values of BD, BC and CD

2x + 7 = 5x - 5 + x

Collect Like Terms

2x - 5x - x = -5 - 7

-4x = -12

Divide both sides by -4

\frac{-4x}{-4} = \frac{-12}{-4}

x = 3

To determine the length of CD;

Substitute 3 for x in CD = x

Hence;

CD = 3

3 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
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