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stira [4]
3 years ago
15

If f(x)= 2(3x²-2x+2). Find:A) f(x+1)B) f(3)​C) f(y-2)

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
8 0
A) f(x+1) = 6x^2 + 8x + 6
1) substitute the value of x as (x+1)
2) 2(3(x+1)^2 -2(x+1) +2)
3) 6(x^2 + 2x + 1)
4) 6x^2 + 8x + 6

B) f(3) = 46
1) simplify the expression first
2) 6x^2 -4x + 4
3) Now just substitute 3: 6(3)^2 -4(3) + 4
4) 6•9 - 12 + 4 = 46

I hope this helps :)
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What is (4/9-7/15) divided by 13/10
Rufina [12.5K]
First find least common denominator to subtract.
9=(3)×3
15=(3)×5
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6 0
3 years ago
Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation edit
Nana76 [90]

Answer:

Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k

As we want a unit vector, we must calculate the modulus of u:

|u|=\sqrt{12^2+0^2+12^2} = \sqrt{2\cdot 12^2}=12\sqrt{2}.

Thus, the wanted vector is \frac{u}{12\sqrt{2}}. Therefore,

u = \frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k = \frac{\sqrt{2}}{2}i-\frac{\sqrt{2}}{2}k.

C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)

where \lambda is a real number.

8 0
4 years ago
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