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lana [24]
3 years ago
5

Pythagorean Theorem in Three Dimensions

Mathematics
2 answers:
exis [7]3 years ago
4 0

Answer:

A is

✔ an edge

.

B is

✔ a vertex

.

C is

✔ a face

.

Step-by-step explanation:

gregori [183]3 years ago
3 0

Answer:

Step-by-step explanation:

I think 3 5 6

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Helga [31]

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15/30 0r 1/2

Step-by-step explanation:

You multiply the fractions

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I Need The Answer Plz!!
hram777 [196]

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9.0

Step-by-step explanation:

In a triangle, the midline joining the midpoints of two sides is parallel to the third side and half as long, so

3x-2 = 50/2

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6 0
4 years ago
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Convert these improper fractions to mixed numbers 74 over 3 please help
noname [10]

PLEASE MARK AS BRAINLIEST

Answer: 24\frac{2}{3}

Step-by-step explanation:

See file attached

3 0
3 years ago
Need help with trig problem in pic
Sidana [21]

Answer:

a) cos(\alpha)=-\frac{3}{5}\\

b)  \sin(\beta)= \frac{\sqrt{3} }{2}

c) \frac{4+3\sqrt{3} }{10}\\

d)  \alpha\approx 53.1^o

Step-by-step explanation:

a) The problem tells us that angle \alpha is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

cos(\alpha)=-\frac{3}{5}

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find sin(\alpha -\beta) by using the identity:

sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\

Notice that we need to find sin(\alpha), which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}

Then:

sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}

d)

Since sin(\alpha)=\frac{4}{5}

then  \alpha=arcsin(\frac{4}{5} )\approx 53.1^o

4 0
3 years ago
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