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son4ous [18]
4 years ago
12

What is the slant height, sss, of the triangular prism?

Mathematics
2 answers:
Andre45 [30]4 years ago
8 0

Answer:

The distance from the middle of the base width to the top peak of the triangle is the base depth. The actual lengths of the triangle's edges that run from the base to the top peak of the triangle represent the slant height.

Step-by-step explanation:

Dafna11 [192]4 years ago
5 0

Answer:

The distance from the middle of the base width to the top peak of the triangle is the base depth. The actual lengths of the triangle's edges that run from the base to the top peak of the triangle represent the slant height.

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The magnitude and direction of two vectors are shown in the diagram. What is the magnitude of their sum? ​
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Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that

<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>

<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>

where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.

Then the sum is

<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>

and its magnitude is

||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)

… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))

… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))

… = √(16 + 16 cos(135° - 45°) + 4)

… = √(20 + 16 cos(90°))

… = √20 = 2√5

5 0
3 years ago
Read 2 more answers
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