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3 years ago

Plz I need your help! Question in picture!

Mathematics

1 answer:

xxMikexx [17]3 years ago

8 0

Answer:

This may be incorrect but $4

Step-by-step explanation:

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Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

7 0

3 years ago

Y+1=3(x−4) what is x and y

kykrilka [37]

Answer:

y=−13

These are the x

and y intercepts of the equation y+1=3(x−4)

x-intercept: (133,0)

y-intercept: (0,−13)

Step-by-step explanation:

4 0

4 years ago

Can I get help please?

irina1246 [14]

Answer:

Round to a number that is not a decimal.

4 0

3 years ago

Morre Math HARDDDDDDDD

Dafna11 [192]

Answer:

the person above is right it's 4.5 units

give them the crown

7 0

3 years ago

I need help <br> 5=6(1q-5)-19

Rudiy27

ok distribute the 6 by multiplying it by 1q and -5
5=6q-30-19
Now subtract 30-19
5=6q-11
add 11 to both sides
16=6q
divide both sides by 6
q= 16/6
simplify
16/2=8, 6/2=3
q=8/3

7 0

3 years ago

Read 2 more answers