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3 years ago

Mary sells lemonade at $1.50 a glass. She spent $8.00 on supplies. She ended with more than $28.00

Mathematics

2 answers:

inna [77]3 years ago

4 0

Answer:

(C.) 1.50n - 8.00 > 28.00

Step-by-step explanation:

Mary sells lemonade at $1.50 a glass

She spent $8.00 on supplies

She ended with more than $28.00

The cost of n lemonade glasses - the cost of supplies is more than $28.00

i.e $1.50n - $8.00 > $28.00

Elena-2011 [213]3 years ago

3 0

Answer:

(C.) 1.50n – 8.00 > 28.00

Hope this helped you!!!

Love,

Morgan Taylor!

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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

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

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

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2 years ago

What is 913 rounded to the nearest hundred

mixer [17]

If i mean .913 then .91 but if u wrote it correctly then 900

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3 years ago

Help, i just need the answers and no need for explanations.

maria [59]

Answer:

The equation of the quadratic function shown is;

x^2+ 2x -3

Step-by-step explanation:

Here in this question, we need to know the quadratic equation whose graph was shown.

The key to answering this lies in knowing the roots of the equation.

The roots of the equation are the solution to the quadratic equation and can be seen from the graph at the point where the quadratic equation crosses the x-axis.

The graph crosses the x-axis at two points.

These are at the points x = -3 and x = 1

So what we have are;

x + 3 and x -1

Multiplying both will give us the quadratic equation we are looking for.

(x + 3)(x-1) = x(x -1) + 3(x-1)

= x^2 -x + 3x -3 = x^2 + 2x -3

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Answer:

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