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Molodets [167]
3 years ago
11

Let u be an angle in the 4th quadrant with cos u=8/9 Then sin u=

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0
To solve for the last side of the triangle, use the Pythagorean Theorem:
(8)^2 + x^2 = (9)^2
x = sqrt of 17
However, this is a NEGATIVE sqrt 17 because the terminal side is in quadrant 4, meaning that this side is under the X-axis and therefore negative.
Now that you know the side opposite of u in the triangle, do opposite/hypotenuse.
sin u = -(sqrt 17)/9
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We solve 6 + 4

6 + 4 = 10.

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\boldsymbol { \sqrt{6 + 4} =  \sqrt{10}  }\\&#10;\\&#10;\\&#10;\boldsymbol {  \sqrt{10}l\underline{ \ 3.16227766017}}\\&#10;\boldsymbol {  \underline{- 10}l \ 3.16227766017 \times 3.16227766017 }\\&#10;\boldsymbol { \ \ \ 00}

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</span>
\boxed{\textbf{3.16227766017} \boldsymbol{ \ \Leftarrow }\textbf{Square root of 10}}

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3 years ago
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ser-zykov [4K]
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3 years ago
The base of a right pyramid is a regular hexagon with sides of length 12 m. The altitude is 6 m. Find the total surface area of
Eva8 [605]

Answer:

806.12 m

Step-by-step explanation:

<u>Step 1: Write the data</u>

<em>Side (s) = 12</em>

<em>Altitude (h) = 6</em>

<u>Step 2: Write the formula for surface area of right hexagonal pyramid</u>

<em>Area = 3√3/2*s² + 3*s*√(h²+3s²/4)</em>

<u>Step 3: Substitute values of 's' and 'h'</u>

<em>Area = 3√3/2*(12)² + 3*(12)*√[6²+3(12)²/4]</em>

Area = 806.12 m

Therefore, the total surface area of the pyramid is 806.12 m.

!!

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Ivanshal [37]
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3 years ago
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