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weqwewe [10]
3 years ago
11

Someone help me with this I will make you brain

Mathematics
2 answers:
kogti [31]3 years ago
7 0

Answer:

\sqrt{5}

Step-by-step explanation:

Working with 3/4 ·2/3 = 6/12 = 1/2

(5^3/4)^2/3 = 5^1/2 = √5

Anettt [7]3 years ago
4 0

Step-by-step explanation:

when having an exponent of an exponent, the exponents are simply multiplied.

so, here we have 3/4 × 2/3 = (3×2)/(4×3) = 6/12 = 1/2

so, simplified, we have 5^(1/2).

now, a fraction as exponent indicates a root to be taken. which root is identified by the denominator of the fraction.

here this is 2 (of 1/2). so we need to take the square root.

therefore

\sqrt{5}

is the right answer.

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Find the indicated coefficients of the power series solution about x=0 of the differential equation
nasty-shy [4]
Write the ODE as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\sum_{n\ge0}a_nx^n=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}x^{2n}

and truncate as many terms as needed to obtain only the terms up to order 4:

(2a_2+6a_3x+12a_4x^2+20a_5x^3+30a_6x^4)-\left(x-\dfrac16x^3\right)(a_0+a_1x+a_2x^2+a_3x^3)=1-\dfrac12x^2+\dfrac1{24}x^4
2a_2+(6a_3-a_0)x+(12a_4-a_1)x^2+\left(20a_5+\dfrac16a_0-a_2\right)x^3+\left(30a_6-a_3+\dfrac16a_1\right)x^4=1-\dfrac12x^2+\dfrac1{24}x^4

We only care about the coefficients up to a_4, so we take the system

\begin{cases}2a_2=1\\6a_3-a_0=0\\12a_4-a_1=-\frac12\end{cases}

Now, we're given initial values y(0)=-3 and y'(0)=7, so that

y(0)=\displaystyle a_0+\sum_{n\ge1}a_n0^n=a_0=-3
y'(0)=\displaystyle a_1+\sum_{n\ge2}na_n0^{n-1}=a_1=7

which gives

a_0=-3,a_1=7,a_2=\dfrac12,a_3=-\dfrac12,a_4=\dfrac{13}{24}
3 0
4 years ago
F the endpoints AB of have the coordinates A(9, 8) and B(-1, -2), what is the midpoint of AB?
lys-0071 [83]
In these kind of problems, it's always the best to just draw it. There are "formulas" to calculate this, but it's better to know intuitively.


Midpoint, as the name suggests is the point that is exactly halfway between two points. To find a midpoint in one dimension, we would take an average between those two points.
Example: Number exactly halfway between  8 and 16 is  \frac{8 + 16}{2}= 12

We can do the same for two dimensions, it just means we'll have to calculate both for x-values and y-values

Midpoint ( \frac{x_{1}+x_{2} }{2} , \frac{y_{1}+y_{2} }{2} ) = ( \frac{9-1 }{2} , \frac{8-2 }{2} ) =( 4, 3 )

6 0
3 years ago
Please HELP
krok68 [10]
1 meter = 100 centimeters
2.5 meters = 250 cm
If you use 50 cm, then you have
2 meters or 200 cm left
8 0
3 years ago
Read 2 more answers
How to find the asymptotes of a hyperbola?
stellarik [79]
Vertical asymptote:
Find the restriction on x. This is the easiest of the three asymptotes you will need to find (even if only two can show at a time). As a hyperbola is in the form: \frac{1}{x}, you only need to find the restriction on the denominator, namely denominator can never be zero. Hence, let the denominator equal to zero to find the vertical asymptote.

Horizontal asymptote:
Find the restriction on y. To do this, you need to simplify the top and bottom to its lowest terms. If it simplifies to a form such as: a + \frac{b}{x}, then the horizontal asymptote becomes y = a. You need to think to yourself, as x grows to infinite, and shrinks to negative infinite, what happens to the function? Does it slowly curve to a stop?

Oblique asymptote:
This is a pretty rare kind, but it still exists, so don't be naive to this sort of asymptote. This is a form of horizontal and vertical asymptote, only it's at an angle. That is, this asymptote is a set of x and y-coordinates that work in unison to produce a curvature or line.

Let's consider: f(x) = \frac{x^{2} - 6x + 7}{x + 5}

Now, in normal term, a horizontal asymptote would have a degree higher in the denominator than in the numerator. However, it's flipped in this case.

Now, you will need to long divide this set of polynomials to yield a straight y = x line, except it's been moved 11 units to the right to yield a y = x - 11 line.

Remember: these exist because the highest power is in the numerator and not the denominator.
5 0
3 years ago
SEE ATTACHMENT BELOW - 40 POINTS
andrey2020 [161]

Answer: C

Step-by-step explanation:

I got it right

C

4 0
3 years ago
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