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3 years ago

I WILL MAKE BRAINIEST!!!!

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

4 0

17 degrees, you just add 22.1 to -5.1

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Decide which title best describes all of the expressions in each column of the table. Place the correct title at the top of each

irga5000 [103]

For the first one is has greater then 3 terms
The middle one is has exactly one term
And the last one is has two terms
I believe I hope this helps

4 0

3 years ago

Material for a dress costs £5.50 per metre. find the cost of 10cm and 70cm

Zielflug [23.3K]

Answer:

The cost of 10 cm and 70 cm are £0.55 and £3.85 respectively.

Step-by-step explanation:

Material for a dress costs £5.50 per meter.

We know that, 1 meter = 100 cm.

Suppose, the cost of 10 cm material is $x$ and the cost of 70 cm material is $y$

Now, according to the ratio of "cost" to the "length of material" , we will get.......

$\frac{x}{10} =\frac{5.50}{100}\\ \\ 100x=10*5.50=55\\ \\ x=\frac{55}{100}=0.55$

and

$\frac{y}{70} =\frac{5.50}{100}\\ \\ 100y=70*5.50=385\\ \\ x=\frac{385}{100}=3.85$

So, the cost of 10 cm material is £0.55 and the cost of 70 cm material is £3.85

5 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

3 years ago

Mental heath day GET SOMETHING OFF YOUR CHEST! (20 points)

Valentin [98]

Answer:

I'm tried of smiling for others and holding all their problems

6 0

3 years ago

Read 2 more answers

Please help with this I will mark brainliest :)

bixtya [17]

Answer:

Triangles

Step-by-step explanation:

Angle b
Triangle BED
Triangle ABC

5 0

3 years ago