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schepotkina [342]
2 years ago
14

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

6 0
3 years ago
Help me solve this paper mathematical
Murrr4er [49]
I wish i can help you
7 0
3 years ago
A total of 676 tickets were sold for the school play. They were either adult tickets or student tickets. There were 74 fewer stu
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!
6 0
2 years ago
PLEASE HELP ASAP, I NEED A PERFECT SCORE. I'M IN A STRESS AND PANIC MODE BECAUSE I MAY NOT FINISH 9TH GRADE, PLZ HELP ME ANSWER
grandymaker [24]
M(x) = 5x + 4     n(x) = 6x - 9

Part 1 (m + n)(x) = 5x + 4 + 6x - 9
                           = 11x - 5

Part 2 (m * n)(x) = (5x + 4)(6x - 9) = 30x^2 - 21x - 36

Part 3 m[n(x)] = 5(6x - 9) + 4
                       = 30x - 45 + 4
            = 30x - 41
5 0
3 years ago
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