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3 years ago

Can someone help me PLS :)

Mathematics

1 answer:

Elodia [21]3 years ago

8 0

Answer:

Step-by-step explanation:

The graph in this case represents exponential decline.

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The ratio of boys to girls in band class was 3 to 5. If there were 24 boys. How many girls were in band class?

Tamiku [17]

Answer:

40 girls

Step-by-step explanation:

Set up a proportion where x is the number of girls in band class:

$\frac{3}{5}$ = $\frac{24}{x}$

Cross multiply and solve for x:

3x = 120

x = 40

So, there were 40 girls in band class.

7 0

2 years ago

Read 2 more answers

Malkia is thinking of a mystery number. She cuts it in half then adds 5. The result is 24. Determine the mystery number. I need

White raven [17]

Answer:38

Step-by-step explanation:if you take 24 and subtract the five she added, it gives you 19 and since she divided it by two you would multiply it by two to get the original number she was thinking of

Hope this helps :)

6 0

3 years ago

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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

2 years ago

Putt the fraction from grates to least value 1/2 5/6 3/4 1/3 1/6 1/4 2/3<br>

gavmur [86]

1. 5/6
2. 3/4
3. 2/3
4. 1/6
5. 1/4
6. 1/2

7 0

2 years ago

Read 2 more answers

There are 2 cups in 1 pint.

Alex17521 [72]

The equation that shows the relationship between the number of cups and pints is c = 2p.

<h3>What is an equation?</h3>

An equation is an expression used to show the relationship between two or more variables and numbers.

Let c represent the number of cups and p represent the number of pints. There are 2 cups in 1 pint. Hence:

c = 2p

When p = 2; c = 2(2) = 4

When p = 3; c = 2(3) = 6

The equation that shows the relationship between the number of cups and pints is c = 2p.

Find out more on equation at: brainly.com/question/2972832

8 0

2 years ago