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2 years ago

8

Write an equation in point-slope form of a line with a slope of –4 that goes through the point J(–5, 6).

1 answer:

Fittoniya [83]2 years ago

5 0

The equation is y = -4x + 26.

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There are not asymptotes for the parent function<br> f (x) =1/x <br> True<br> False

Georgia [21]

Answer:

True

Step-by-step explanation:

the parent function f(x) = 1/x has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0

4 0

2 years ago

99 plus 99b gyzuosfk

Anettt [7]

Answer:

hmm...

Step-by-step explanation:

3 0

2 years ago

2nq=3+7nq solve for n thanks

Gnesinka [82]

Answer:

n = - $\frac{3}{5q}$

Step-by-step explanation:

Given

2nq = 3 + 7nq ( subtract 7nq from both sides )

- 5nq = 3 ( divide both sides by the multiplier - 5q )

n = $\frac{3}{-5q}$ = - $\frac{3}{5q}$

7 0

2 years ago

Read 2 more answers

65 points Of the 30 girls who tried out for the basketball team at SMS, 12 were

natulia [17]

Answer:

the ratio is the same

Step-by-step explanation:

Girls

12 out of 30

12/30

Divide top and bottom by 6

2/5

Boys

16 out of 40

16/40

Divide top and bottom by 8

2/5

8 0

2 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

2 years ago