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crimeas [40]
2 years ago
11

A small radio transmitter broadcasts in a 29 mile radius. If you drive along a straight line from a city 32 miles north of the t

ransmitter to a second city 40 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter
Mathematics
1 answer:
hram777 [196]2 years ago
7 0

Answer: He will pick up a signal from the transmitter 29.43 miles of the drive.

Step-by-step explanation:

Since we have given that

Radius = 29 mile

So, equation of circle would be

x^2+y^2=29^2=841

There is a straight line from a city 32 miles north i.e. (0,32) to a second city 40 miles east of the transmitter i.e. (40,0)

So, equation of line would be

y-32=\dfrac{0-32}{40-0}(x-0)\\\\y-32=\dfrac{-32}{40}x\\\\y-32=-0.8x\\\\y+0.8x=32\\\\y=32-0.8x

So, we will put this value in the equation of circle.

x^2+(32-0.8x)^2=841\\\\x^2+1024+0.64x^2-51.2x=841\\\\1.64x^2-51.2x=841-1024\\\\1.64x^2-51.2x=-183\\\\1.64x^2-51.2x+183=0

So, x= 4.12 and x = 27.1

So, y = 28.7, y = 10.32

So, distance between them would be

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(27.1-4.12)^2+(28.7-10.32)^2}\\\\=29.43

Hence, he will pick up a signal from the transmitter 29.43 miles of the drive.

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a)

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b)

his income must be "x" which is the 100%, and we would know that 2% of that is 12.37.

\begin{array}{ccll} amount&\%\\ \cline{1-2} x&100\\ 12.37&2 \end{array}\implies \cfrac{x}{12.37}=\cfrac{100}{2}\implies 2x=1237 \\\\\\ x=\cfrac{1237}{2}\implies x=618.5

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What are the roots to<br> the equation<br> 2x² + 6x-1=0?
lys-0071 [83]

Answer:

The roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}    

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = \frac{ - b}{a}

And products of roots = \frac{c}{a}

So, m + n = \frac{ - 6}{2} = - 3

And m × n =  \frac{ - 1}{2}

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 (\frac{ - 1}{2})

Or, (m - n)² = 9 + 2 = 11

I.e m - n = \sqrt{11}

Again m + n = - 3    And m - n = \sqrt{11}

Solving this two equation

(m + n) + ( m - n) = - 3 + \sqrt{11}

I.e 2 m =  - 3 + \sqrt{11}

Or, m = \frac{-3}{2} + \frac{\sqrt{11} }{2}

Similarly n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      

Hence the roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      Answer

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