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Citrus2011 [14]
3 years ago
12

If A=(3,2) and B=(5,7) be two points , find the unit vector along BA

Mathematics
1 answer:
Alexeev081 [22]3 years ago
5 0
If we were to plot the points on a graph, we could draw a straight lint between them. The question wants us to find the length of this line. We could make a right triangle with this line as the hypotenuse, with an x length of 2(y2-y1) and a y length of 5 (y2-y1). From there we can find the hypotenuse with Pythagorean Theorem: c^2=a^2+b^2. 
c^2=(2)^2+(5)^2
c^2=4+25
C^2=41
c=\sqrt{41}
Decimal approximation: 6.403

Hope this helps!
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Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

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The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

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The equation of the quadratic function.

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A quadratic function is defined as:

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It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

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31=a(5)^2+b(5)+11

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Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

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0=3(2)+b

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Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

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