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Vilka [71]
3 years ago
9

A circular cake is cut into 8 equal pieces. What is the angle measure of each piece?

Mathematics
2 answers:
pishuonlain [190]3 years ago
4 0

Answer:

45

Step-by-step explanation:

A whole circle is 360 so if you divided that by 8 you get 45

kakasveta [241]3 years ago
4 0

Answer: 45°

Step-by-step explanation:

The angles of a circle have to add up to 360°

Since the pieces have to be equal, you have to divide 360 by 8.

360 ÷ 8 = 45

The angle measure of each piece is 45°  

Hope I helped! (°;ω;°)

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When 2(y^2) + 8 is divided by  2y + 4 is equal to (y - 2) + (16 / (2y + 4)). The expression represents the quotient is the 2y + 4. While the expression represent the remainder is 16 / (2y + 4). The remainder of the given expression can also be solve using the remainder theorem. 

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Find the exact values of all three trigonometric functions for each angle.
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Answer:

Sin (270º)= -1

Cos (270º) = 0

Tan (270º) = -∞

Sin (330º) = -0.5

Cos (330º) = \frac{\sqrt{3} }{2} = 0.8660

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Step-by-step explanation:

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Answer:

First, subtract 3 from both sides.

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x>6

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2 years ago
A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t
solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by  

V = V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x

First, we differentiate V(x) with respect to x, to get,

\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450

Equating the first derivative to zero, we get,

\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0

Solving, with the help of quadratic formula, we get,

x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2},

Again differentiation V(x), with respect to x, we get,

\frac{d^2(V(x))}{dx^2} = 24x - 180

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\displaystyle\frac{5(3-\sqrt{3})}{2},

\frac{d^2(V(x))}{dx^2} < 0

Thus, by double derivative test, the maxima occurs at

x = \displaystyle\frac{5(3-\sqrt{3})}{2} for V(x).

Thus, largest volume the box can have occurs when x = \displaystyle\frac{5(3-\sqrt{3})}{2}}.

Maximum volume =

V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}

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3 years ago
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answer: 16

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