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2 years ago

The radius of Earth is 6,378.1 km, which is 18,983.9 km less than the radius of Uranus. Find the radius of Uranus.

Mathematics

1 answer:

Anton [14]2 years ago

4 0

Uranus has an equatorial radius of 25,559 km and a radius of 24,973 km at the poles.

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I need the answer asap pls! :)

AfilCa [17]

Answer:

18m

Step-by-step explanation:

Since the given traingles are similar

We get

$\frac{JK}{MN} = \frac{KL}{ML}$

$\frac{JK}{42} = \frac{21}{49}$

$JK = \frac{21 \times 42}{49}$

$JK = \frac{3 \times 7 \times 6 \times 7}{7 \times 7}$

$JK = 3 \times 6 = 18$

$JK = 18$

Hence the distance across river is 18m

7 0

1 year ago

A dress is on sale for 15% discount, meaning 85% of the price remains. If the sale price of the dress is $51, then the original

ira [324]

Answer:

Step-by-step explanation:

let the original price=x

85 % of x=51

$x=51*\frac{100}{85} =60$

5 0

2 years ago

A fair six-sided dice is rolled once. What is the probability that a 2 is shown?

VashaNatasha [74]

Answer:

C: 1/6

Step-by-step explanation:

There is one 2 on a six sided die. Therefore there is a 1/6 chance that it was a 2.

3 0

3 years ago

Read 2 more answers

From the set (33, 15, 12), use substitution to determine which value of x makes the equation true.

zepelin [54]

X=12 because 3 times 12 equals 36. if you divide by 3 on both sides you would be left with that

8 0

3 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

2 years ago