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TiliK225 [7]
3 years ago
12

Solve for x. 2(4x - 1) + 3(3 - 2x) = 0 x = -4 x = -7/2 x = -4/3 x = -7/6

Mathematics
2 answers:
lara [203]3 years ago
7 0
The correct answer: x=-7/2
Hope this help!!!
romanna [79]3 years ago
5 0

I'm positive that it is -7/2

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I need details please
m_a_m_a [10]
Ok, first you need to read a - g and mark down the information you know. Also, this continuous function can look pretty much like anything long as it has the a - g constraints/rule. 

A) We know the domain(x values) is between -10 to infinity
(-10,-9,-8,-7,-6,-5,-4,-3-,2,-1,0,1,2,3,4,5,6,7,8,9,10, to ---------> infinity)
B) We know the range(y values) is between -6 to infinity
(-6,-5,-4,-3-,2,-1,01,2,3,4,5,6,7,8,9,10, to ---------> infinity) 

C) Next we know f(0) = f(4) = 0
So we know when x = 0  that f(0) = 0 or we can say  y = 0. We also know when f(4) = 0 so when x = 4 y = 0. So we have two points (0,0) and (4,0) that our function will be on so plot the two points (0,0) (4,0)

D) It tells us that f(x) (y) is increasing between the x values (-10,-6)  or (-2,2) or (4, to infinity)

E) It tells us that f(x) (y) is decreasing between the x values (-6,-2) or between (2,4)

F) Tells us that f(x) (y) is greater than or equal to between the x values of [-8,-4] or between the x values [0, to infinity) so this lets us know our drawing is going to be above or equal to 0 when we are in the values [-8,-4] or [0, to infinity)

G) Tells us that f(x) (y) is less than 0 when the function / drawing is between the x values of (-10, -8) or (-4,0)  

The picture is what the function could look like. It can be different but needs to following the rules that are given 




3 0
3 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

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Answer:  The standard deviation was calculated incorrectly.

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It cannot be negative , because it is the square root of the variance .

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⇒ Standard deviation cannot be negative.

⇒The standard deviation was calculated incorrectly.

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