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2 years ago

Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).

Mathematics

1 answer:

strojnjashka [21]2 years ago

5 0

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Daniel dives into a swimming pool. The height of his head is represented by the function h(t) = -8t2 - 28t + 60, where t is the

8_murik_8 [283]

So, h(t) is how far is Daniel's head from the surface of the water, namely the surface itself is when the height is nill, so his head is at the surface and the height of it is just 0, thus h(t) = 0. Namely, what is "t" when h(t) is 0?

$\bf h(t)=-8t^2-28t+60\implies \stackrel{h(t)}{0}=-8t^2-28t+60
\\\\\\
0=-2t^2-7t+15\implies 2t^2+7t-15=0\implies (2t+3)(t-5)=0
\\\\\\
\begin{cases}
2t+3=0\implies 2t=-3\implies &t=-\frac{3}{2}\\\\
t-5=0\implies &\boxed{t=5}
\end{cases}$

clearly the seconds cannot be a negative unit, so is not -3/2.

6 0

3 years ago

What is the slope of line PQ?

Anit [1.1K]

Answer:

Step-by-step explanation:

Slope of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

Slope = $\frac{y_2-y_1}{x_2-x_1}$

Slope of the line PQ passing through the two points P(-8, 2) and Q(4, 2) will be,

Slope ' $m_1$ ' = $\frac{2-2}{-8-2}$ = 0

Therefore, slope of the line PQ parallel to the x-axis = 0

Slope of the line MN passing through the points M(8, 6) and N(8, -8) is,

Slope ' $m_2$ ' = $\frac{6+8}{8-8}$ = ∞

Therefore, slope of the line MN parallel to y-axis is undefined.

Since, angle between the x-axis and y-axis is 90°,angle at the intersection point between the lines parallel to x and y axis will be 90°.

These lines are perpendicular to each other.

4 0

3 years ago

A.16cm b. 16 root 2cm c. 8 root 2cm d. 8cm

Stolb23 [73]

Answer:

Step-by-step explanation:

Mark as Brainliest :)

7 0

3 years 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

What is the value of x? 750 37°

RSB [31]

Whats the problem? i cant solve it w/o the problem doll

3 0

2 years ago