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3 years ago

Explain why the denominator does not change when you are counting by the

Mathematics

1 answer:

VMariaS [17]3 years ago

5 0

Answer:

Yes, when dividing a whole number by a unit fraction, multiplying the whole number by the unit fraction's denominator always works! a divided by (1/b) = a times (b/1) = (a/1) times (b/1) = ab/1 = ab. Have a blessed, wonderful day! Comment on Ian Pulizzotto's post “Yes, when dividing a whole number by a unit fracti...”

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find h and k such that the equation 3x^2 -hx +4k =0, so that the sum of the solutions is -12 and the product of the solutions is

Radda [10]

Let the solutions be a and b.
a = -2; b = -10
a + b = -2 + (-10) = -12
ab = (-2)(-10) = 20

(x - a)(x - b) = 0

(x - (-2))(x - (-10)) = 0

(x + 2)(x + 10) = 0

x^2 + 10x + 2x + 20 = 0

x^2 + 12x + 20 = 0

-h = 12

h = -12

4k = 20

k = 5

6 0

3 years ago

What’s the answer? Down below is a image of the question

kotykmax [81]

Answer:

Step-by-step explanation:

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3 0

2 years ago

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Donna bought 5 bags of dog treats for $14.00. What is the cost per bag of dog treats?

Alex17521 [72]

Answer:

2.8

Step-by-step explanation:

14 divided by 5=2.8

Hope this helps you! <3

6 0

2 years ago

Read 2 more answers

What is the basic ratio for 30:42

lianna [129]

30:42 is represented as 30/42
The lowest form for that is 5/7. Hope that helped. Good luck

6 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

2 years ago