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Lelu [443]
3 years ago
14

What are the statements and reason that make these two triangles similar?

Mathematics
1 answer:
pochemuha3 years ago
8 0

the 2x is the same side

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The reflection of a Quadrant II point is located in Quadrant I. Assuming no error was made, what kind of reflection occurred?
gizmo_the_mogwai [7]

Answer:

A reflection across the y- axis

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y) → (-x, y)

A point in the second quadrant (- x, y)

Under reflection in the y- axis is (x, y) ← point in first quadrant

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Marco is carrying two cylinder-shaped cans. Each can has the same height
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Choice D

Step-by-step explanation:

The Cavalieri's principle states that if two or more solids have identical cross-sectional area as well as height, then the solids will have equal volume.

Therefore, the volumes of the cylinders are the same

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You are ordering shirts for the math club at your school. Short-sleeved shirts cost $10 each. Long-sleeved shirts cost $12 each.
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Answer:

10x+12y=300

Step-by-step explanation:

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Helpppppppp pls help .
dsp73

Answer: X= 1 2 3 4  Y= 5 8 11  14

Step-by-step explanation: i tried my best

5 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%28%20%5Csin%5E%7B2%7D%20%28%20%5Cfrac%7B%5Cpi%7D%7B%204%20%7D%20%20-%20%20%5Calpha%20%29%20%2
guapka [62]

Step-by-step explanation:

\sin^2 (\frac{\pi}{4} - \alpha) = \frac{1}{2}(1 - \sin 2\alpha)

Use the identity

\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}

on the left side.

\dfrac{1 - \cos [2(\frac{\pi}{4} - \alpha)]}{2} = \frac{1}{2}(1 - \sin 2\alpha)

\dfrac{1 - \cos (\frac{\pi}{2} - 2\alpha)}{2} = \frac{1}{2}(1 - \sin 2\alpha)

Now use the identity

\sin \theta = \cos(\frac{\pi}{2} - \theta)

on the left side.

\dfrac{1 - \sin 2\alpha}{2} = \frac{1}{2}(1 - \sin 2\alpha)

\frac{1}{2}(1 - \sin 2\alpha) = \frac{1}{2}(1 - \sin 2\alpha)

4 0
3 years ago
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