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Anna11 [10]
3 years ago
15

3x+2y=6 and y=-3/2x+5

Mathematics
1 answer:
love history [14]3 years ago
6 0
In order to solve for parallel, perpendicular, or neither, you have to look at the slope.

If the slope is the same for both equations, it is most likely parallel.
If it's the reciprocal (Where you flip the number and add change the signs. For example, the reciprocal of 1/2 is -2)
If the slope is not the same or the reciprocal, then it is neither. 

So for the first equation, your slope is:
3x+2y=6
2y=-3x+6
y=-3/2x+3      The equation y=mx+b can help you here, where m is the slope.
Your slope is -3/2

For the second equation, your slope is -3/2 since y=-3/2x+5 is already in y=mx+b form and m is the slope.

Since both slopes are -3/2, then you have parallel equations!

(Be careful though, sometimes it will have the same slope but there will also be the same y-intercept. If that happens, it's no longer parallel, but it's the same equation. Such as y=-3/2x+1 and y=-3/2x+1. In this case there will be infinite solutions, but parallel equations have no solutions.)

I hope this helps!! Please ask if you have more questions!
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Step-by-step explanation:

find the coordinate of the foot of the perpendicular from(4,-2) to the line 2x-3y-4=0

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3 years ago
Helppppp please i need to know the answer right now! Thanks!
Ivenika [448]

Answer:

f(7) - f(-7)

= (4 - 7^2) - ((-7)^3 - 2*(-7))

= -45 - (-329)

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3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, −1), B(4, −3, 0), C(1
zavuch27 [327]

Answer:

Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

Step-by-step explanation:

1. First find the length of the vectors  

AB = B-A = (4-1, 3-0, 0-1) = (3, 3, -1)

AC = C-A = (1-1, 4-0, 3-1) = (0, 4, 2)

BC = C-B = (1-4, 4-3, 3-0) = (-3, 1, 3)

2. Find magnitude of the vectors

|AB| = √(3^2+3^2+〖(-1)〗^2 )=√19  

|AC| = √(0^2+4^2+2^2 )=2√5

|BC| = √(〖(-3)〗^2+1^2+3^2 )=√19

3. Find the angles between them

cos θ = (a.b)/(|a||b|) -----> θ = arc cos ((a.b)/(|a||b|))

AB, AC : θ =arc cos (AB.AC)/(|AB||AC|) = arc cos ( (3.0+3.4+ -1.2)/(√19  x 2√5)=  10/(2√95) ) =59.136, which is approximately 59.14°

AB, BC : θ =arc cos (AB.BC)/(|AB||BC|) = arc cos( (3.-3+3.1+ -1.3)/(√19  x √19)=  (-9)/19 ) = 118.27°

Because of the direction of BC is pointing relative to AB this is the angle outside the triangle, and we should find the supplementary angle.

180-118.27 = 61.73°

If we used –(BC) = CB in the formula (just negate the numerator) we would have gotten the correct angle on first try.

The 3rd angle should be what’s left after subtracting from 180°;

180-61.73-59.14 = 59.13 °

You can confirm using the formula again :

AC, BC : θ =arc cos (AC.BC)/(|AC||BC|) = arc cos( (0.-3+4.1+ 2.3)/(√19  x 2√5)=  10/(2√95) = 59.13°

Rounding everything up so they add up to 180 degrees we have Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

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The answer would be $2.70

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