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avanturin [10]
3 years ago
11

A right triangle has a leg of 12 cm and a hypotenuse of 16 cm what is the length of the other leg

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
The length of the other leg is 12 cm
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The representative sample contained more girls than boys.

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Which face has an area of 28 cm
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trapezium?

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Thirteen is at least the difference of a number v and 1.
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13≥v-1 is the formula
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A rectangular room is 11 feet by 18 feet. The ceilings are 10 feet high. If a gallon of paint covers 250 square feet. How many g
Bingel [31]

Answer:

2.32 gallons of paint

Step-by-step explanation:

First, find the area of all 4 sides of the room, with the equation A = lw

11(10)

= 110 (for 2 of the walls)

18(10)

= 180 (for the other 2 walls)

Add these up:

110 + 110 + 180 + 180

= 580

Then, divide this by 250:

580/250

= 2.32

So, 2.32 gallons of paint will be needed to paint the room.

5 0
3 years ago
Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me​
o-na [289]

Answer:

\frac{3}{2}

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

                   = - cos60cosx - sin60sinx

                   = - \frac{1}{2} cosx - \frac{\sqrt{3} }{2} sinx

squaring to obtain cos² (120 + x)

= \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

                   = -cos60cosx + sin60sinx

                   = - \frac{1}{2}cosx + \frac{\sqrt{3} }{2}sinx

squaring to obtain cos²(120 - x)

= \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x + \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

= cos²x + \frac{1}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}(cos²x + sin²x) = \frac{3}{2}

                 

5 0
3 years ago
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