A small radio transmitter broadcasts in a 31 mile radius. If you drive along a straight line from a city 38 miles north of the transmitter to a second city 36 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter? Hint: assume that the transmitter is at the origin. Write the equations of the line and circle.

9514 1404 393

  63.7% or 33.35 miles

Step-by-step explanation:

The hint suggests that we find the points where the drive begins and ends picking up the transmitter. We suspect both of those points will have irrational coordinates, so finding the distance between them will involve dealing with squares and roots of irrational numbers. We want to see if there's an easier way.

The distance from the transmitter to the path being driven can be found using the "distance to a line" formula. First, we need the equation of the line in general form. In intercept form, it is ...

  y/38 +x/36 = 1

  19x +18y -684 = 0 . . . . . . multiply by LCM(36, 38) and subtract that amount

Then the distance from the origin to the line is ...

  d = |19·0 +18·0 -684|/√(19² +18²) = 685/√685 ≈ 26.1243

The distance (x) along the line from the point where the transmitter is first picked up until the point of closest approach can be found from the Pythagorean theorem:

  x² + (26.1243)² = 31²

  x² = 31² -(684/√685)² = 190429/685

  x ≈  16.6733

The driving distance for which the transmitter is picked up is twice this, so is ...

  2x = 2(16.6733) ≈ 33.3466 . . . miles

The total drive length is also given by the Pythagorean theorem:

  drive length = √(36² +38²) = √2740 ≈ 52.3450 . . . miles

Then the fraction of the drive during which the signal is picked up is ...

  fraction = (33.3466 mi)/(52.3450 mi) ≈ 0.6371 ≈ 63.7%

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The signal is picked up for 33.35 miles, about 63.7% of the drive.

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The intercept-form equation for a line is ...

  x/(x-intercept) +y/(y-intercept) = 1

The distance from (x, y) to line ax+by+c=0 is given by ...

  d = |ax+by+c|/√(a²+b²)

The Pythagorean theorem relates legs a, b and hypotenuse c of a right triangle this way:

  c² = a² +b²

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )