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enot [183]
3 years ago
14

Write the equation of a line in - form that contains the point (- 1, - 2) and is perpendicular to the line y = 5x - 10 degrees

Mathematics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

y+2= -1/5(x+1)

Step-by-step explanation:

if lines are perpendicular their slopes are negative reciprocal

y=mx+b where m is the slope

y=5x-10 has the slope 5 so a perpendicular line will have slope -1/5

equation point slope form

(y-y1) = m(x-x1) where m is slope, and (x1,y1) any point that belongs to the line

y+2= -1/5(x+1)

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Edgar Anderson earns $200 a week plus a 15% commission on all sales over $1,000. If Mr. Anderson's sales for one week are $2,500
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3 0
4 years ago
Read 2 more answers
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
3 years ago
Read 2 more answers
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