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3 years ago

Simplify : 3a(2a-x+1) URGENTLYYY

Mathematics

1 answer:

Delicious77 [7]3 years ago

8 0

Answer:

the answer is 6a^2- 3ax+ 3a

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Given 11 different natural numbers, none greater than 20. Prove that two of these can be chosen, one of which divides the other.

frosja888 [35]

Answer:11 honestly have no idea but im going to give my best answer it is 31

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3 years ago

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Find the length of the third side. If necessary, round to the nearest tenth.

koban [17]

Answer:

7.5

Step-by-step explanation:

Using pythagoras theorem

Sqrt(9^2-5^2)=7.483

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2 years ago

write the equation of a line that is perpendicular to the given line and that passes through the given point. y-4=5/2(x+3) ; (-7

faltersainse [42]

The gradient of the perpendicular line would be the negative reciprocal of the original line. Therefore the gradient of the perpendicular line would be -2/5x.
Since we know y=mx+c, ∴y=-2/5x+c. Sub in the x and y values of the given point and we get that c=26/5.

The perpendicular equation would be y=-2/5x+26/5.
I hope I got this right.

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3 years ago

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Selena was comparing the flying speeds of different birds in science class. Which of the three birds flies fastest? Why?

Mashcka [7]

Answer:

Option C. The robin flies fastest because its rate is 15 m/s

Step-by-step explanation:

Let

x -----> the time in seconds

y ----> the distance in meters

The speed is equal to divide the distance by the time

In this problem, the slope or unit rate of the linear equation is the same that the speed

Robin's Flight

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

take two points from the table

(2,30) and (3,45)

substitute in the formula

$m=\frac{45-30}{3-2}$

$m=\frac{15}{1}$

$m=15\frac{m}{sec}$

Cardinal's Flight

$y=10x$

$m=10\frac{m}{sec}$

Blue Jay's Flight

take two points from the graph

(0,0) and (2,20)

substitute in the formula

$m=\frac{20-0}{2-0}$

$m=\frac{20}{2}$

$m=10\frac{m}{sec}$

Compare

The robin flies fastest because its rate is 15 m/s

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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

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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2 years ago