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3 years ago

13

Solve: -18 - 5y > 52

2 answers:

geniusboy [140]3 years ago

5 0

Answer:

y < -14.

Step-by-step explanation:

-18 - 5y > 52

-5y > 52 + 18

-5y > 70

y < 70/-5 (Note the inequality sign flips when dividing by a negative).

y < -14

exis [7]3 years ago

3 0

Answer:

y<-14

Step-by-step explanation:

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3 years ago

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4 years ago

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3 years ago

A) Write the following in standard form and find A, B and C

alex41 [277]

Answer:

a) 2x -y = -5; A=2, B=-1, C=-5
b) y = 2x +5

Step-by-step explanation:

a) Standard Form is ...

Ax + By = C

where A > 0 and A, B, C are mutually prime.

To get that form, we can add 2x-5 to both sides of the equation:

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

-5 = 2x -y

Then we can swap sides of the equal sign:

2x -y = -5 . . . . . . standard form

Matching this to the generic standard form equation, we find ...

A = 2

B = -1

C = -5

__

b) We can solve for y by adding 2x+y to both sides of the equation:

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

y = 2x +5 . . . . . . slope-intercept form

5 0

3 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

So

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago