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Gennadij [26K]
3 years ago
8

To finish a project, you need to paint the lateral surfaces of a cube with side length 2.5 inches. Find the area that you need t

o paint.
Mathematics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

25 in^2

Step-by-step explanation:

We want to find the lateral surface area of the cube.

The lateral surface area of a cube is given as:

A = 4a^2

where a = length of side of the cube

Therefore, given that the length of the side of the cube is 2.5 inches, the area you need to paint is:

A = 4 * 2.5^2\\A = 25 in^2

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IceJOKER [234]

Answer:

111°

Step-by-step explanation:

  • All these are parallel lines, so the 36° angle is equal to the 36° angle inside the big triangle because they are vertically opposite.

  • Ignore the line cutting between 45° and the 30° and consider it as one triangle
  • Add them to get 75°
  • Now you have two known angles 75° and 36°
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3 years ago
A military cannon is placed at the base of a hill. The cannon is fired at an angle toward the hill. The path
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Answer:

Algebracicaly speaking the answer would be either -13.3876 or - 158.612 through the quadratic equation, but these answers don’t make sense in this real world scenario.

Step-by-step explanation:

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2 years ago
(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.
valentinak56 [21]

Find the intersection of the two planes. Do this by solving for <em>z</em> in terms of <em>x</em> and <em>y </em>; then solve for <em>y</em> in terms of <em>x</em> ; then again for <em>z</em> but only in terms of <em>x</em>.

-4<em>x</em> + 2<em>y</em> - <em>z</em> = 1   ==>   <em>z</em> = -4<em>x</em> + 2<em>y</em> - 1

3<em>x</em> - 2<em>y</em> + 2<em>z</em> = 1   ==>   <em>z</em> = (1 - 3<em>x</em> + 2<em>y</em>)/2

==>   -4<em>x</em> + 2<em>y</em> - 1 = (1 - 3<em>x</em> + 2<em>y</em>)/2

==>   -8<em>x</em> + 4<em>y</em> - 2 = 1 - 3<em>x</em> + 2<em>y</em>

==>   -5<em>x</em> + 2<em>y</em> = 3

==>   <em>y</em> = (3 + 5<em>x</em>)/2

==>   <em>z</em> = -4<em>x</em> + 2 (3 + 5<em>x</em>)/2 - 1 = <em>x</em> + 2

So if we take <em>x</em> = <em>t</em>, the line of intersection is parameterized by

<em>r</em><em>(t)</em> = ⟨<em>t</em>, (3 + 5<em>t</em> )/2, 2 + <em>t</em>⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

<em>r</em><em>(t)</em> = ⟨2<em>t</em>, 3 + 5<em>t</em>, 4 + 2<em>t</em>⟩

(a) The tangent vector to <em>r</em><em>(t)</em> is parallel to this line, so you can use

<em>v</em> = d<em>r</em>/d<em>t</em> = d/d<em>t</em> ⟨2<em>t</em>, 3 + 5<em>t</em>, 4 + 2<em>t</em>⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

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<em>r</em><em>(t)</em> = ⟨2<em>t</em>, 3 + 5<em>t</em>, 4 + 2<em>t</em>⟩

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The slope-intercept form: y = mx + b

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Answer:32

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