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Natasha2012 [34]
2 years ago
6

Use Pythagorean theorem to find the missing length. (round to tenth if necessary) IS

Mathematics
1 answer:
sertanlavr [38]2 years ago
7 0

11.7

Formula: a^2 + b^2 = c^2

so substitute the numbers like this:

10^2 + 6^2 = c^2

both 10^2 and 6^2 are perfect squares:

10 × 10= 100 and 6 × 6= 36

100 + 36= c^2

add:

100 + 36 =

136 = c^2

now in order to get rid of the squared, you put it into a radical:

√136 = √c^2

note: whatever you do on one side, you do to the other

it's easier to use the calculator for this so put √136 and you're gonna get 11.66

since it's asking you to round it to the nearest tenth:

11.66 = 11.7

sorry if the explanation is kinda long ✋ i tried explaining it better in the end but i failed

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Answer: fluffy = 33yrs

Spot = 19 yrs

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Step-by-step explanation:

Let fluffy = x

Spot = y

Skampy =z

x + y + z = 91 -------1

x -14 =y ------2

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Put eqn 2 and 3 in eqn 1

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3 years ago
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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
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Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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The green rectangle has area 200*50 = 10000 square meters. The green semicircles combine to form an area of pi*r^2 = pi*25^2 = 625pi square meters. In total, the full green area is 10000+625pi square meters. I'm leaving things in terms of pi for now. The approximation will come later.

The blue area is the same story, but smaller dimensions. The blue rectangle has dimensions 200 meters by 42 meters, so its area is 200*42 = 8400 square meters. The blue semicircular pieces combine to a circle with area pi*r^2 = pi*21^2 = 441pi square meters. In total, the blue region has area 8400+441pi square meters.

After we figure out the green and blue areas, we subtract to get the orange region's area, which is the area of the track only.

orange area = (green) - (blue)

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