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3 years ago

How long does it take to travel a distance of 672 km at a speed of 96 kph?

Mathematics

2 answers:

mote1985 [20]3 years ago

4 0

Answer:

7 hrs

Step-by-step explanation:

formula :- speed=distance/time

speed= 96kph

distance= 672 km

time=?

S=D/T

96=672/T

(cross multiply)

96T = 672

T = 672/96

T = 7 hrs

mr_godi [17]3 years ago

3 0

<h2>Answer:</h2>

It takes 7 hours to travel a distance of 672 km at a speed of 96 kph.

<h2>Step-by-step explanation:</h2><h3>Known :</h3>

Speed = 96kph
Distance = 672km

<h3>Asked :</h3>

Time

<h3>Solution :</h3>

Time = Distance/speed

Time = 672 / 96

Time = 7

<h3>Conclusion :</h3>

It takes 7 hours to travel a distance of 672 km at a speed of 96 kph.

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If cos(θ)=2853 with θin Q IV, what is sin(θ)?

marysya [2.9K]

Answer: $\sin \theta=\frac{-45}{53}$

Step-by-step explanation:

Since we have given that

$\cos\theta=\frac{28}{53}$

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

$\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}$

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7 0

3 years ago

A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

salantis [7]

Dimensions are length 20 meter and width 14 meter

<em><u>Solution:</u></em>

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

<em><u>A rectangle has width that is 6 meters less than the length</u></em>

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

<em><u>The area of the rectangle is given by formula:</u></em>

$Area = length \times width$

<em><u>Substituting the values we get,</u></em>

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

<em><u>Solve the above equation by quadratic formula</u></em>

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$

Since, length cannot be negative, ignore a = -14

<em><u>Thus solution of length is a = 20</u></em>

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

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3 years ago