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3 years ago

How would you scale down by a factor of 1/10?

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

Answer:

See example below.

Step-by-step explanation:

To scale down by a factor of 1/10, multiply the side length or distance by 1/10.

Example:

If the distance was 5 then 5*1/10 = 5/10 = 1/2 is the scaled down version.

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Write your answers as mixed numbers.

max2010maxim [7]

19/2=9 1/2

9 1/2 bottles

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2 years ago

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For his exercise last weekend, Benito walked 5 miles and jogged 2 miles. What is the ratio of the miles walked to the total numb

Sergeu [11.5K]

Answer:

ummm pepepopo

Step-by-step explanation:

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3 years ago

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Matthew is using a microscope in science class. The microscope magnified the specimen 4x10^2 . If the specimen is 3cm long how l

choli [55]

Answer:

12 meters

Step-by-step explanation:

The apparent size is ...

(4·10^2)·(3 cm) = 12·10^2 cm = 12 m

___

1 m = 10^2 cm

5 0

3 years ago

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

In Exercise,find the horizontal asymptote of the graph of the function.<br> f(x) = 8x^3+2/2x^3+x

KIM [24]

Answer:

Horizontal asymptote of the graph of the function f(x) = (8x^3+2)/(2x^3+x) is at y=4

Step-by-step explanation:

I attached the graph of the function.

Graphically, it can be seen that the horizontal asymptote of the graph of the function is at y=4. There is also a <em>vertical </em>asymptote at x=0

When denominator's degree (3) is the same as the nominator's degree (3) then the horizontal asymptote is at (numerator's leading coefficient (8) divided by denominator's lading coefficient (2)) $y=\frac{8}{2}=4$

6 0

3 years ago