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Ilia_Sergeevich [38]
3 years ago
15

What is the axis of symmetry for y= x2 - 6x + 2

Mathematics
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

the axis of symmetry is 3.

Step-by-step explanation:

the axis of symmetry is the line that passes through the vertex.

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The price of 4 dresses is $486 What is the rate of each dress?
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D. is the correct answer your just divide 486 by 4 you get 121.50
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What is the constant of variation for the quadratic variation? x 2 3 4 5 6 y 32 72 128 200 288
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We use different models for different types of variation. For example, linear variation is associated with the formula y=ax, or the more familiar y=mx+b (the equation of a straight line). Cubic variation: y=a*x^3. In the present case we're discussing quadratic variation; perhaps that will ring a bell with you, reminding you that y=ax^2+bx+c is the general quadratic function. Now in y our math problem, we're told that this is a case of quadratic variation. Use the model y=a*x^2. For example, we know that if x=2, y =32. Mind substituting those two values into y=a*x^2 and solving for y? Then you could re-write y=a*x^2 substituting this value for a. Then check thisd value by substituting x=3, y=72, and see whether the resulting equation is true or not. If it is, your a value is correct. But overall I got 16!
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Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your
BaLLatris [955]

f(x,y)=\dfrac{y^3}x

a. The gradient is

\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath

\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}

b. The gradient at point P(1, 2) is

\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}

c. The derivative of f at P in the direction of \vec u is

D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}

It looks like

\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath

so that

\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2

Then

D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}

\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}

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3 years ago
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