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BARSIC [14]
3 years ago
15

What is the slope of the line conaning the points (-3,4)and(2,1)

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -3 &,& 4~) 
%  (c,d)
&&(~ 2 &,& 1~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-4}{2-(-3)}\implies \cfrac{1-4}{2+3}\implies \cfrac{-3}{5}
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3 years ago
You have a summer job that pays time and a half for overtime. that is if you work more than 40 hours per week,your hourly wage f
Nataliya [291]
Your pay is described by a piecewise function p(t) where t is hours worked.
p(t)=\left\{\begin{array}{rcl}7t&\mbox{for}&t\leq40\\10.5t-140&\mbox{for}&t>40\end{array}\right.

The domain of this function is the number of hours you can work in a week—zero to 168. The range is the set of values p(t) can take on, which we have said is p(0) to p(168). This problem, however, is asking for the range for a typical 40-hour week. That will be p(0) to p(40): the interval [0, 280].
8 0
3 years ago
D square=3h+s<br>change of subject formula<br>make h the leader<br>​
Ksju [112]

Answer: h=(D²-s)/3

Step-by-step explanation: D² =3h+s

Subtract a from both sides of the equation

D²-s =3h+s-s

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D²-s =3h

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(D²-s)/3 =3h/3

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(D²-s)/3 =h

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7 0
2 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}}

4 0
2 years ago
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