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3 years ago

Captain Ishaan has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Luis and his merciless

Mathematics

1 answer:

fredd [130]3 years ago

4 0

Answer:

Khan Academy's answer is 4/35.

Step-by-step explanation:

4/5 * 1/7 = 4/35

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mary is preparing for her college entrance exams. in a practice test, she answered 12 problems in 30 minutes. at this rate, how

mariarad [96]

12 problems = 30 minutes
24 problems = 60 minutes
36 problems= 90 minutes
48 problems = 120 minutes
60 problems = 150 minutes

Mary can respond 60 problems in 2 hours and 30 minutes/ 2 hours and half/ 150 minutes.

Hope this helps xx

3 0

3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Victoria and Georgetown are 67.2 miles from each other. How far apart would the cities be on a map that has a scale of 0.9 inche

Artemon [7]

Answer: 5.76 miles

Step-by-step explanation:

Let the distance between Victoria and Georgetown on the map be represented by x.

The information given in the question can then be formed into an expression which will be:

0.9/x = 10.5/67.2

Cross multiply

(10.5 × x) = 67.2 × 0.9

10.5x = 60.48

x = 60.48/10.5

x = 5.76

The answer is 5.76 miles

7 0

3 years ago

Find the distance between the following coordinates: I(-a,b), J(a,b)

olchik [2.2K]

Answer:

Step-by-step explanation:

using distance formula:

√(x₂-x₁)² + (y₂-y₁)²

putting values,

√{a-(a)}² + {b-b}² = √ {a+a}² + 0²

= √(2a)² = √4a² = 2a

Hope this helps:)

8 0

3 years ago

The scale of a map is 6 km = 4 mi. What is the length on the map if the actual is 40.1<br> miles.

beks73 [17]

Answer:

60.15 km

Step-by-step explanation:

If 6 km on the map represents 4 mi on land, then we'd get the answer by using the relation

6 km -> 4 mi

x km -> 40.1 mi

Where x km is the length on the map we're looking for. If we cross multiply

6 * 40.1 = 4x

240.6 = 4x

x = 240.6/4

x = 60.15 km.

That is the length we're looking for

8 0

3 years ago