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Vlad1618 [11]
3 years ago
13

X^2 = 5

Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0
I think the answer is D
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How to solve the 4a.plz
vladimir1956 [14]
(4xy-2y^2)+2y
4-2=2....x will remain y^1-y^2=y^-1
2xy^-1+2y
answer=2x+2y
3 0
2 years ago
Which of the graphs below would result if you made the leading term of the
Anettt [7]

Answer:

A%7.2= ( 7.5№) =(+--+) this is the best answer

3 0
3 years ago
Read 2 more answers
Solve N/3 + (-4)=-2.<br> A. 6<br> B. -18<br> C. 18<br> D. -2
Nana76 [90]

Answer:

A) 6

Step-by-step explanation:

n/3+(-4)=-2

n/3-4=-2

n/3=-2+4

n/3=2

n=2*3

n=6

6 0
2 years ago
b) If parametric equations of a flow line are x = x(t), y = y(t), explain why these functions satisfy the differential equations
sineoko [7]

Answer:

The equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1.

Step-by-step explanation:

The pathline equation for a vector field is given by F(x,y) = xî - yj

The velocity vector field for the streamline of the flow is given by

V(x, y) = (dx/dt)î + (dy/dt)j

From the question, it is given that

(dx/dt) = x

(dy/dt) = -y

Hence, the velocity vector field for the streamline of the flow in question is

V(x, y) = xî - yj

which coincides with the pathline vector field of the flow.

The only time the pathline and streamline vector field coincide and have the same equation is when the flow is a steady state flow.

That is, the properties of the fluid flowing isn't changing with time!

Hence, this flow is a steady state flow!

We're told to solve the differential equation.

(dx/dt) = x

(dy/dt) = -y

but

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = -y/x

(dy/y) = -(dx/x)

∫(dy/y) = -∫ (dx/x)

In y = - In x + c

where c is the constant of integration

In y + In x = c

In (xy) = c

Inserting the values of (x, y) given in the question,

In (-1 × -1) = c

In 1 = c

0 = c

c = 0

In y + In x = 0

In (yx) = 0

xy = e⁰ = 1

xy = 1

So, the equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1

Hope this Helps!!!

4 0
3 years ago
To mix plaster for a dental model, 35 mL of water is used for 100g of plaster. How many mL of water should be used for 250g of p
laila [671]

Okay, here we have this:

Considering the provided information, we are going to calculate the requested value, so we obtain the following rule of three:

\begin{gathered} \frac{x}{35mL\text{ of water}}=\frac{250g\text{ of plaster}}{100g\text{ of plaster}} \\ x=\frac{250g\text{ of plaster}}{100g\text{ of plaster}}\cdot35mL\text{ of water} \end{gathered}

Solving for x:

\begin{gathered} x=2.5\cdot35mL\text{ of water} \\ x=87.5mL\text{ of water} \end{gathered}

Finally we obtain that 87.5mL of water should be used for 250g of plaster.

3 0
9 months ago
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