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Wewaii [24]
2 years ago
15

Lyle graphs triangle ABC on coordinate axes. He performs two transformations on this figure that result in the congruent triangl

e XYZ Which series of transformations did Lyle most likely use?
a translation left and then a dilation with a scale factor of 3
a dilation with a scale factor of 2.5 and then a rotation 180' counterclockwise
a rotation 90' clockwise and then a reflection across the y-axis
a reflection across the x-axis and then a dilation with a scale factor of 0.25
C​
Mathematics
1 answer:
Shtirlitz [24]2 years ago
3 0

Answer:

a rotation 90˚ clockwise and then a reflection across the y-axis

Step-by-step explanation:

I'm doing the test right now

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What is the area? (IXL)
Zina [86]

Answer:

45 square meters

Step-by-step explanation:

we know that

The area of trapezoid is equal to

A=\frac{1}{2}(b1+b2)h

where

b1 and b2 are the parallel bases

h is the height of trapezoid (perpendicular distance between the parallel bases)

we have

b1=6\ m\\b2=9\ m\\h=6\ m

substitute

A=\frac{1}{2}(6+9)6

A=45\ m^2

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Step-by-step explanation:

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Step-by-step explanation:

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A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th
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The boat went 5hours upstream, let's say it has a "still water" speed rate of "b", it went 100miles... however, going upstream is going against the current, let's say the current has a speed rate of "c"

so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down  "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

now, recall your d = rt, distance = rate * time

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
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\end{cases}

solve for "c", to see what's the current's speed

what's "b"?  well 20+c = b
4 0
3 years ago
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