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3 years ago

What are the steps to solve/verify this

Mathematics

2 answers:

vlabodo [156]3 years ago

8 0

Hope this helps
pls ask if you don't understand

Troyanec [42]3 years ago

5 0

Tan^2 x - Tan^ x sin^2 x

= tan^2x ( 1 - sin^2 x)

= tan^2 x * cos^x

= sin^2 x * cos^2 x
----------------------
cos^2 x

= sin^2 x

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Christine wants to buy a bicycle for her little brother. She receives a coupon in the mail for 5% off the bicycle. The bicycle h

puteri [66]

The answer is A because you would do 79 x 0.05 which equals 3.95 . So you subtract 3.95 from 79 which is 75.05 minus 50 equal to 25.05

5 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

Please help me with questions 6

loris [4]

Since A = bh
and A = 198 and b = 18?
h = A/b
= 198/18
= 11

6 0

3 years ago

NEED HELP ON THIS QUESTION

Tom [10]

Answer: 288

Workings : 16 x 12 =192 (for bottom rectangle bit) 12x 16 =192 then divide 192 by 2 =96 add them together and its 288

3 0

3 years ago

If you graph 2y+8=6x^2-10x what will be the y intercept?

Rufina [12.5K]

Answer:

Y intercept will be -4.

Step-by-step explanation:

Here we are given our function as

$2y+8=6x^2-10x$

We are asked to determine the y intercept.

The y intercept is the y ordinate of the coordinates of the point at which the graph of the function intersects the y axis.

Please note that the point at which graph cuts the y axis have x=0

Hence in order to determine y intercept we need to put x=0 in our function and solve it for y :

$2y+8=6x^2-10x2y+8=6x^2-10x$

Putting x =0

$2y+8=6x^2-10x2y+8=6(0)^2-10(0)$

$2y+8=6x^2-10x2y+8=0-10(0)$

$2y+8=6x^2-10x2y+8=0$

subtracting both sides by 8

$2y+8=6x^2-10x2y=-8$

dividing both sides by 2

=-4

8 0

4 years ago