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PtichkaEL [24]
3 years ago
11

Two less than the product of 8 and a number is 8

Mathematics
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

x=1.25

Step-by-step explanation:

Let us replace the number with the variable "x". You will then get this equation:

8x-2=8

Move the -2 to 8 and you would get the equation:

8x=8+2

Simplify:

8x=10

Simplify:

x=10/8

You would then get the answer "x=1.25"

I hope this helps :)

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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
How do I solve this question ??????
Vlada [557]
Since you have 4 people total splitting the money, you can find the total amount made in the two months by: 184(4)=736
Since you want the total profit in the second month, subtract the total amount by the profit in the first month: 736-438=298
So the answer is $298
3 0
3 years ago
Choose the correct word that completes each statement about inscribing a square in a circle.
Anna11 [10]

Answer:

1. Perpendicular

2. Isosceles

3. Never

Step-by-step explanation:

1. AC ⊥ BD because diameter of a square are perpendicular bisector of each other.

2. In Δ AOB , By using pythagoras : AB² = OA² + OB² .......( 1 )

In Δ COB , By using pythagoras : BC² = OC² + OB²  ..........( 2 )

But, OA = OC because both are radius of same circle

So, by using equations ( 1 ) and ( 2 ), We get AB = BC ≠ AC

⇒ ABC is a triangle having two equal sides so ABC is an isosceles triangle.

3. The side can never be equal to radius of circle because the side of the square will be chord for the circle and in a circle chord can never be equal to its radius


7 0
3 years ago
Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the
mafiozo [28]

Answer:

Porcentage of students score below 62 is close to 0,08%

Step-by-step explanation:

The rule

68-95-99.7

establishes:

The intervals:

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ] contains 68.3 % of all the values of the population

[ μ₀ - σ ,  μ₀ + σ]   contains 95.4 % of all the values of the population

[ μ₀ - 1,5σ ,  μ₀ + 1,5σ] contains 99.7 % of all the values of the population

In our case such intervals become

[ μ₀ - 0,5σ ,  μ₀ + 0,5σ]   ⇒  [ 78 - (0,5)*8 , 78 + (0,5)*8 ]  ⇒[ 74 , 82]

[ μ₀ - σ ,  μ₀ + σ]  ⇒ [ 78 - 8 , 78 +8 ]   ⇒  [ 70 , 86 ]

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]  ⇒ [ 78 - 12 , 78 + 12 ]  ⇒ [ 66 , 90 ]

Therefore the last interval

[ μ₀ - 1,5σ ,  μ₀ + 1,5 σ]    ⇒  [ 66 , 90 ]

has as lower limit 66 and contains 99.7 % of population, according to that the porcentage of students score below 62 is very small, minor than 0,15 %

100 - 99,7  = 0,3 %

Only 0,3 % of population is out of   μ₀ ± 1,5 σ, and by symmetry 0,3 /2 = 0,15 % is below the lower limit, 62 is even far from 66 so we can estimate, that the porcentage of students score below 62 is under 0,08 %

6 0
3 years ago
4n+12=4 What does n equal?
solong [7]

Answer:

n = -2

Step-by-step explanation:

4n + 12 = 4

Subtract 12 from both sides.

4n = -8

Divide both sides by 4.

n = -2

3 0
3 years ago
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