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Paha777 [63]
3 years ago
15

When a figure is translated, reflected, or rotated, what's the same about the original igure and its image? What's different?

Mathematics
1 answer:
Liula [17]3 years ago
7 0

Answer:

The image of a translation, reflection, or rotation is congruent to the original figure, and the image of a dilation is similar to the original figure. Two figures are similar when one can be obtained from the other by a sequence of translations, reflections, rotations, and dilations.

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A local University accepted 2,300 students out of 4,500 applicants for admission. What was the acceptance rate, expressed as a p
Mademuasel [1]
2300 out of 4500 were chosen

2300 / 4500 = 0.5111

0.51 = 51%

51% is the acceptance rate.
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Svetllana [295]
I wish I could help but but I can't I'm dumb as a box of rocks
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Can 3 and 3/5 also be rewritten as 3 and 15/20?
mario62 [17]
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1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

and computing it is trivial:

\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

7 0
3 years ago
The average cost of tuition and fees at a public four-year college in the United States can be molded by the following formula w
Rudik [331]
T=7556=165x+2771
we need to solve for x, number of years after 1996.
165x=7556-2771=4785
x=4785/165=29
It will be in year 1996+29=2025 that estimated fees and tuition average $7556
3 0
3 years ago
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