All categories

3 years ago

2+1/6y=3x+4 for y can someone please help

Mathematics

2 answers:

Mashutka [201]3 years ago

4 0

It would be y=18x+12

Maksim231197 [3]3 years ago

3 0

Answer:
y=18x+12

Explanation:
2+1/6y=3x+4
Subtract both sides by 2
2+1/6y-2=3x+4-2
1/6y=3x+2
Multiply both sides by 6 (to get rid of the denominator)
1/6y*6=3x*6+2*6
y=18x+12

I hope this helps!

You might be interested in

Find the discriminant and the number of real roots for the equation. 4x^2+12x+9=0

netineya [11]

Answer:

Step-by-step explanation:

5 0

3 years ago

Rewrite the quadratic y=3(x-2)^2-7 in standard form

DochEvi [55]

Answer:

The answer is

y=3/(x-2)^5

8 0

3 years ago

Solve the following initial-value problem: (ex+y)dx+(3+x+yey)dy=0, y(0)=1 0=

alex41 [277]

Answer:

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

Step-by-step explanation:

Given that

$(e^x+y)dx+(3+x+ye^y)dy=0$

Here

$M=e^x+y$

$N=3+x+ye^y$

We know that

M dx + N dy=0 will be exact if

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y}=1$

$\frac{\partial N}{\partial x}=1$

it means that this is a exact equation.

$\int d\left(e^x+xy+3y+(y-1)e^y\right)=0$

Noe by integrating above equation

$e^x+xy+3y+(y-1)e^y=C$

Given that

x= 0 then y= 1

$e^0+0+3+(1-1)e^1=C$

C=4

So the our final equation will be

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

3 0

3 years ago

Will give me a 20% discount if I spend over $50 my bill is $75.What is my final bill after the savings

Andrews [41]

Hey there,
$75 is above $50
Final Bill = $75 - (20% x $75)
= $75 - $15
= $ 60

Hope this helps :))

~Top♥

7 0

3 years ago

Read 2 more answers

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago