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viktelen [127]
2 years ago
7

A part of Jennifer’s work is to solve the equation 2(6x^2-3)=11x^2-x is shown below

Mathematics
1 answer:
kotykmax [81]2 years ago
7 0

Answer:

Step-by-step explanation:

2(6x² - 3) = 11x² - x

2*6x² - 2*3 = 11x² - x

12x² - 6 = 11x² -x

Subtract 11x² from both sides

12x² - 11x² - 6 = -x

x² - 6 = -x

x² + x - 6= 0

Sum =1

Product = -6

Factors =  3 , (-2)              { 3*(-2)  = -6   & 3 +(-2) = 1}

x² + 3x - 2x - 6 = 0

x(x + 3) - 2(x + 3)= 0

(x +3)(x - 2) = 0

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Which ordered pair is a solution of the equation ?
Gekata [30.6K]

Answer:

C. Both of the above answers

Step-by-step explanation:

13 = -6(-2) + 1

13 = 12 + 1

13 = 13

7 = -6(-1) + 1

7 = 6 + 1

7 = 7

7 0
3 years ago
Identify the vertex and the axis of symmetry of the graph of the function y=3(x+2)2-3
dedylja [7]
Axis -b/2a=-12/2*3=-2
vertex(-2,-3)
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3 years ago
Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)
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g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}

The surface element is

\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

and the integral is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)

###

To compute the last integral, you can integrate by parts:

u=r\implies\mathrm du=\mathrm dr

\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}

\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr

For this integral, consider a substitution of

r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds

\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds

\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds

=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds

and the result above follows.

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3 years ago
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3 years ago
Hey uhm can someone answer this for 12 points
MAXImum [283]

\frac{0.2x - 3.2}{5}  = 1.8 \\ 0.2x - 3.2 = 1.8 \times 5 \\ 0.2x - 3.2 = 9 \\ 0.2x = 9 + 3.2 \\ 0.2x = 12.2 \\ x =  \frac{12.2}{0.2}  \\ \boxed{ x = 61}

4 0
2 years ago
Read 2 more answers
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