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2 years ago

Convert: 90 kilograms to pounds 198.4 1,984 19,840 198,400

Mathematics

2 answers:

Paha777 [63]2 years ago

7 0

90 kilograms is 198.4 pounds

lana [24]2 years ago

5 0

Answer:

198.4

Step-by-step explanation:

TO convert kilograms to pounds you just have to make a simple multiplication, remember that there are 2.204 pound in each kilogram, so we just do a rule of thirds:

$\frac{1kg}{2.204 pounds} =\frac{90}{x}$

$x=\frac{(90)(2.204)}{1}$

$x=198.4$

SO the answer is 90 kilograms are equal to 198.4 pounds

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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

Aiden and Haisem are going to eat the same amount of hamburgers and fries, but from different restaurants. Aiden goes to Burger

Paha777 [63]

Answer:

the burgers be 3 and fries be 2

Step-by-step explanation:

The computation is shown below:

Let us assume burgers be x

And, the fries be y

Now according to the questiojn

1.25x + 0.50y = $4.75

1.50x + 0.99y = $6.48

Now multiply by 1.2 in the first equation

1.50x + 0.6y = $5.70

1.50x + 0.99y = $6.48

-0.39y = -0.78

y = 2

Now put the value of y in any of the above equation

1.25x + 0.50(2) = $4.75

x = 3

Hence, the burgers be 3 and fries be 2

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3 years ago

Help me solve this paper mathematical

Murrr4er [49]

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3 years ago

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katen-ka-za [31]

To find the answer,we can solve this by settin up an equation.

Let x be the number of adult tickets sold.

The number of students ticket sold would be:
x-74

We can then sum them up to set an equation:

x+x-74 = 676
2x = 676 + 74
x = 375

Therefore the adult tickets sold were 375.

Hope it helps!

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2 years ago

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grandymaker [24]

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3 years ago