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2 years ago

What does this expression represent five times the quotient of some number and ten

Mathematics

1 answer:

marishachu [46]2 years ago

3 0

Answer:

5(n/10)

Step-by-step explanation:

quotient of some number and ten=n/10

five time=*5

put it togther=5(n/10)

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2. Solve the following equations for 0° Sx < 180°.

diamong [38]

Answer:

See below in bold.

Step-by-step explanation:

(1) cosec ( x + 10) = 3

Now cosec x = 1/sin x so

1 / sin (x + 10) = 3

3 sin (x + 10) = 1

sin (x + 10) = 1/3

x + 10 = 19.47 , 160.53 degrees

x = 9.47, 150.53 degrees.

(ii) cot (x - 30) =0.45

cot (x - 30)= 1 /tan (x- 30) so we have

tan (x - 30) = 1 / 0.45 = 2.2222

x - 30 = 65.77 degrees

x = 95.77 degrees.

5 0

2 years ago

Which is greater, 3/4 or 11/16? How can you be sure?

Aleksandr-060686 [28]

3/4 because if you multiply to get the LCM (16) it becomes 12/16 which is greater than 11/16

4 0

2 years ago

4n/5=84/15<br><br> i cannot figure out this math problem for my homework!

Vinil7 [7]

Okay so instead of dividing you will cross multiply.
4n(15)=84(5)
60n=420
420/60=7
Answer: n=7

4 0

2 years ago

Find the sum of the first 6 terms of the sequence: –3, 1, 5, 9, . . .A) 45B) 42C) 48D) 25

Leni [432]

SOLUTION

From the sequence give

–3, 1, 5, 9, . . .

The first term, a = -3

The common difference, d = 4 (gotten by adding 4 to the next term).

The number of terms required n = 6.

Formula for sum of an arithmetic sequence is given by

$S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack$

Substituting these values into the equation above we have

$\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ S_6=\frac{6}{2}\lbrack2\times-3+(6-1)4\rbrack \\ S_6=3\lbrack-6+(5)4\rbrack \\ S_6=3\lbrack-6+20\rbrack \\ S_6=3\lbrack14\rbrack \\ S_6=42 \end{gathered}$

Hence, the answer is 42, option B

3 0

7 months 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