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3 years ago

What is the decimal expansion of the following fraction? 1/8 A. 0.125 B. 0.125 C. 1.8 D. 0.18

Mathematics

1 answer:

qaws [65]3 years ago

3 0

Answer:

0.125....................

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13. A new bank customer with $5,000 wants to open a money market account. The bank is

9966 [12]

Answer:

a) 900 dollars

b) 5900 dollars

Step-by-step explanation:

The complete question is

A new bank customer with $5,000 wants to open a money market account. The bank is offering a simple interest rate of 1.8%. a. How much interest will the customer earn in 10 years? b. What will the account balance be after 10 years?

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3 years ago

Luke subtracted 128° from 180° and found that the measurement of ∠STR is 52°. How can Luke use this information to find the meas

Karolina [17]

STR is a triangle as shown in the picture. If Luke decided to subtract 128º from 180º, that's because he understood that the sum of the internal angles in a triangle is 180º, and therefore the angles RST and TRS together make 128º.
As a result, RST will be
128º - TRS
where TRS is the angle at R, as shown in the picture

6 0

3 years ago

Read 2 more answers

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

3 years ago

Find the compunt intrest on rs 31250 at 12 % per annum for 12 1/2 years

postnew [5]

Answer:

Step-by-step explanation:

given Principle at 31250

and rate = 12%

and time 12.5 years

the amount would be given by the equation

$P(1+\frac{rate}{100})^{time}$

plug in values to get

$31250(1+\frac{12}{100})^{12.5}$

end up with value $128847

3 0

2 years ago

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter

VMariaS [17]

Answer:

He must survey 123 adults.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.