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Tom [10]
3 years ago
14

The picture below shows a box sliding down a ramp: A right triangle ABC has measure of angle ABC equal to 90 degrees and measure

of angle ACB equal to 65 degrees. The length of AB is 12 feet. What is the distance, in feet, that the box has to travel to move from point A to point C?
12 divided by sec 65 degrees
12 cosec 65°
12 sin 65°
12 divided by cot 65 degrees
Mathematics
2 answers:
Lady_Fox [76]3 years ago
4 0

Answer:

12 cosec 65

Step-by-step explanation:

MrRissso [65]3 years ago
3 0

The Image of the triangle is missing and so i have attached it.

Answer:

B: 12 cosec 65°

Step-by-step explanation:

From the diagram attached, we cans see the triangle is a right angle triangle with a given internal angle of 65° and side AB with a length of 12 ft.

Now, we can use trigonometric ratios to find the distance the box has to move from point A to point C which is basically the distance AC.

Now, the given side AB is opposite to the angle. While the required side AC is the hypotenuse.

Thus;

Opposite/hypotenuse = sin θ

Thus;

AB/AC = 12/AC = sin 65

Thus;

AC = 12/sin 65

From trigonometric relationships, we know that;

1/sin θ = Cosec θ

Thus;

AC = 12 cosec 65

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Goryan [66]

Answer:

Step-by-step explanation:

Let the two numbers be 2x and 3x.

According to the question,

2x+3x=45

5x=45

x=45/5

x=9

so th required numbers are 18 and 27 as

2x=2*9=18

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Answer:

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Step-by-step explanation:

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3 years ago
Reduced form of 2ab^2-a^2 b^2/5
Ket [755]

Answer:

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

Step-by-step explanation:

Given the expression

2ab^2-\frac{a^2b^2}{5}

\mathrm{Convert\:element\:to\:fraction}:\quad \:2ab^2=\frac{2ab^25}{5}

2ab^2-\frac{a^2b^2}{5}=\frac{2ab^2\cdot \:5}{5}-\frac{a^2b^2}{5}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

                 =\frac{2ab^2\cdot \:5-a^2b^2}{5}

                 =\frac{10ab^2-a^2b^2}{5}

Thus,

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

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