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zmey [24]
3 years ago
5

80 POINTS + BRAINLIEST !!!!!!!!

Mathematics
2 answers:
Gre4nikov [31]3 years ago
8 0

Solution:

we are given that

Clayton needs to reflect the triangle below across the line y=x.

And as we know the reflection of the point (x,y) across  the line y = x is the point (y, x).

Hence the new Traingle will be on the other side of the line y=x and position of x and y-coordinates of the vertices of the trangle gets interchanged.

Hence the Options that applies are:

C’ will move because all points move in a reflection.

The image and the pre-image will be congruent triangles. (Because reflection just changes the position of the trinagle not the property)

The image and pre-image will not have the same orientation because reflections flip figures.

GaryK [48]3 years ago
5 0

Point C’ will move because all points move in a reflection.  The image and the pre-image will be congruent triangles

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The ratio of the side lengths of a quadrilateral is 3:3:5:8, and the perimeter is 380cm. What is the measure of the longest side
castortr0y [4]

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160

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Find the integral, using techniques from this or the previous chapter.<br> ∫x(8-x)3/2 dx
Soloha48 [4]

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Step-by-step explanation:

For this case we need to find the following integral:

\int x(8-x)^{3/2}dx

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\int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C

And rewriting in terms of x we got:

\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C

And that would be our final answer.

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3 years ago
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