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3 years ago

If you travel 200 miles in 300 minutes. Howfast did you travel in MPH...mile per hour?

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

5 0

Answer:

40 miles per hour

Step-by-step explanation:

Change 300 minutes to hours. There are 60 minutes to hours

300 minutes * 1 hour/ 60 minutes = 5 hours

Now take the miles and divide by hours

200 miles / 5 hours = 40 miles per hour

VMariaS [17]3 years ago

5 0

Answer: 40 mph

Step-by-step explanation: 300 minutes is 5 hours. 40 mph for 5 hours is 200 miles traveled.

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Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

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1 year ago

Please help me out with this question

I am Lyosha [343]

Answer:

no. 4

Step-by-step explanation:

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3 years ago

Solve by using square root: <br> 4(x-1)^2+2=10

Scilla [17]

$4(x-1)^2+2=10\\\\4(x^2-2x+1)+2-10=0\\\\4 x^2-8x+4+2-10=0\\\\4 x^2-8x-4=0\ \ / :4\\\\x^2-2x-1=0\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{2-\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2-\sqrt{ 8}}{2 }=\frac{2- \sqrt{ 4*2}}{2 }=\\\\=\frac{2-\sqrt{ 8}}{2 }=\frac{2(1- \sqrt{ 2})}{2 }=1- \sqrt{ 2}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2+\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2(1+ \sqrt{ 2})}{2 }=1+ \sqrt{ 2}$

$Answer:\ x=1- \sqrt{2}\ \ or\ \ x= 1+ \sqrt{ 2}$

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3 years ago

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The perimeter of a rectangle is 26 meters. The difference between the length and the width is 5 meters. Find the width

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Try making a substitution:

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