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Brut [27]
3 years ago
8

A cyclist rides a distance of 36 km at an average speed of 16 km/h.

Mathematics
1 answer:
Advocard [28]3 years ago
3 0

Answer:

2.25 hours

Step-by-step explanation:

Given data

distance= 36 km

speed= 16 km/h

Speed= distance/time

substitute

16= 36/t

t= 36/16

t= 2.25 hours

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Answer:

D = \frac{25\sqrt{13}}{13}

Step-by-step explanation:

Given

y = 4 - \frac{2}{3}x

(x_1,y_1) = (-2,-3)

Required

Determine the distance

y = 4 - \frac{2}{3}x

Write the above equation in standard form:

Ax + By + C = 0

So, we have:

\frac{2}{3}x+y  - 4 = 0

By comparison:

A = \frac{2}{3}  B = 1 and C = -4

The distance is calculated using:

D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Where:

(x_1,y_1) = (-2,-3)

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This gives:

D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}

D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}

Take LCM

D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}

D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}

D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}

D = \frac{25}{3}/\sqrt{\frac{13}{9}}

Split the square root

D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}

Change / to *

D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}

D = \frac{25}{3}*\frac{3}{\sqrt{13}}

D = \frac{25}{\sqrt{13}}

Rationalize

D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}

D = \frac{25\sqrt{13}}{13}

Hence, the distance is:

D = \frac{25\sqrt{13}}{13}

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Answer:

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Step-by-step explanation:

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Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle (\frac{1}{3},\frac{1}4).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A(\frac{1}{3},\frac{1}4)

Let O be the origin i.e. (0,0)

Point B will be (\frac{1}{3},0)

Now, let us consider the right angled triangle \triangle OBA:

Sides:

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Using Pythagorean theorem:

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cos\angle AOB = \dfrac{Base}{Hypotenuse}

\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}

tan\angle AOB = \dfrac{Perpendicular}{Base}

\Rightarrow tan\angle AOB = \dfrac{3}{4}

cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}

\Rightarrow cosec\angle AOB = \dfrac{5}{3}

sec\angle AOB = \dfrac{Hypotenuse}{Base}

\Rightarrow sec\angle AOB = \dfrac{5}{4}

cot\angle AOB = \dfrac{Base}{Perpendicular}

\Rightarrow cot\angle AOB = \dfrac{4}{3}

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Step-by-step explanation:

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