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Yuri [45]
4 years ago
9

Rewrite the quadratic function in vertex form. Y=3x^2-12x+4

Mathematics
1 answer:
Karolina [17]4 years ago
5 0

Answer:

vertex form is y=a(x−h)2+k. To solve this you have to complete the square with the x terms: y= 3x2−12x+4. first isolate the x terms: y−4=3x2−12x. ax2+bx+c to complete the square a=1 and c=(12b)2.

Step-by-step explanation:

vertex form is y=a(x−h)2+k. To solve this you have to complete the square with the x terms: y= 3x2−12x+4. first isolate the x terms: y−4=3x2−12x. ax2+bx+c to complete the square a=1 and c=(12b)2.

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topjm [15]

Answer:

Step-by-step explanation:

perp. 4/3

y - 4 = 4/3(x - 3)

y - 4 = 4/3x - 4

y = 4/3x

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3 years ago
Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

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7 0
3 years ago
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???????????I don't know exactly what you mean


6 0
3 years ago
ZE and ZF are vertical angles with mZE = 8x + 8 and mZF = 2x +38.<br> What is the value of x?
Dovator [93]

Answer:

Since ZE and ZF are vertical angles, they are congruent.

8x+8 = 2x+38

8 = 2x-8x+38

8 = -6x+38

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x = -30/-6

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7 0
3 years ago
Read 2 more answers
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serg [7]

Answer:

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6 0
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