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Virty [35]
3 years ago
14

the radius of one circular field is 5m and that of other is 3m.find the radius of circular field whose area is the difference of

the areas of the first and second field.​
Mathematics
1 answer:
Alex777 [14]3 years ago
3 0

Answer:

4m

Step-by-step explanation:

  1. Recall the formula for area of a circle for each calculation
  2. Calculate the area of the large field (r = 5m)
  3. Calculate the area of the small field (r = 3m)
  4. Find the difference in area
  5. Find the radius of the "difference" field using difference in area

The formula for area of a circle is A = πr².

"A" means area.

π is pi. <em>I will use the π button</em>, but some teachers ask to use 3.14.

"r" means radius.

Area of large field:

Substitute "r" for 5m in the formula.

A = πr²

A = π(5m)²

A = π25m²

A = 78.5398163397 m²

Area of small field:

Substitute "r" for 3m in the formula.

A = πr²

A = π(3m)²

A = π9m²

A = 28.2743338823  m²

Difference in areas:

Subtract the area of small field from the area of large field.

(78.5398163397 m²) - (28.2743338823  m²)

= 50.2654825 m²

Radius of "difference" field:

Since we are looking for radius, not area, rearrange the formula to isolate "r".

A = πr²

\frac{A}{\pi } = \frac{\pi r^{2}}{\pi}     Divide both sides by π

\frac{A}{\pi } = r²                π cancels out on the right side

\frac{A}{\pi } = √r²             Square root both sides

\sqrt{\frac{A}{\pi }} = r        ² and √ cancel out, leaving "r" isolated

r = \sqrt{\frac{A}{\pi }}       Variable on the left for standard formatting

Substitute "A" for the difference in area

r = \sqrt{\frac{A}{\pi }}

r = \sqrt{\frac{50.2654825m^{2}}{\pi }}     Divide by pi first

r = \sqrt{16.0000000135&#10;m^{2}}            Find the square root

r ≈ 4m

The radius of the field that has the area of the difference in the two fields is 4m.

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kykrilka [37]

ANSWER

D. 25 feet

EXPLANATION

The height of the wall,h, the taut wire and the distance from the base of the pole to the point on the ground, formed a right triangle.

According to the Pythagoras Theorem, the sum of the length of the squares of the two shorter legs equals the square of the hypotenuse.

Let the hypotenuse ( the length of the ) taught wire be,l.

Then

{l}^{2}  =  {h}^{2}  +  {b}^{2}

{l}^{2}  =  {20}^{2}  +  {15}^{2}

{l}^{2}  =  400 + 225

{l}^{2}  =  625

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8 0
3 years ago
An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

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