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2 years ago

(07.05 LC) How many solutions does the equation 4y + 7 = 5 + 2 + 4y have? One Two None Infinitely many

Mathematics

2 answers:

Elza [17]2 years ago

8 0

Answer: Infinitely many

Step-by-step explanation:

First simplify the right side by adding 5 and 2

now you have the equation 4y+7=7+4y

That is the exact same thing so any number works.

Reil [10]2 years ago

7 0

just to add to the great reply above.

$\bf \begin{matrix} 4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}+7=5+2~~\begin{matrix} +4y \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}\implies 7=7$

whenever you end up with something like 0 = 0, or 7 = 7, is a flag that both equations are exactly the same, in this case, the one on the right-hand-side is really the one one the left-hand-side <u>in disguise</u>.

So the graph of one, is the same as the graph of the other, or put in another words, let's say the graph the first one, the second one when graphed, will just be pancaked on top of the first one, and any point whatsoever on the second one, matches with the first one, and since both lines continue infinitely, then infinitely many solutions.

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For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago

What is the quotient when (x+3) is divided into the polnomial 2x^2+x-15

Alika [10]

Answer:

2x-5

remainder of 0

Step-by-step explanation:

Check the image

7 0

2 years ago

Solve system of equations 2x + 2y + 5z = 7 6x + 8y + 5z = 9 2x + 3y + 5z = 6

svet-max [94.6K]

Answer:

the values of x, y and z are x= 2, y =-1 and z=1

Step-by-step explanation:

We need to solve the following system of equations.

We will use elimination method to solve these equations and find the values of x, y and z.

2x + 2y + 5z = 7 eq(1)

6x + 8y + 5z = 9 eq(2)

2x + 3y + 5z = 6 eq(3)

Subtracting eq(1) and eq(3)

2x + 2y + 5z = 7

2x + 3y + 5z = 6

- - - -

_____________

0 -y + 0 = 1

-y = 1

=> y = -1

Subtracting eq(2) and eq(3)

6x + 8y + 5z = 9

2x + 3y + 5z = 6

- - - -

______________

4x + 5y +0z = 3

4x + 5y = 3 eq(4)

Putting value of y = -1 in equation 4

4x + 5y = 3

4x + 5(-1) = 3

4x -5 = 3

4x = 3+5

4x = 8

x= 8/4

x = 2

Putting value of x=2 and y=-1 in eq(1)

2x + 2y + 5z = 7

2(2) + 2(-1) + 5z = 7

4 -2 + 5z = 7

2 + 5z = 7

5z = 7 -2

5z = 5

z = 5/5

z = 1

So, the values of x, y and z are x= 2, y =-1 and z=1

5 0

3 years ago

A farmer grows corn, tomatoes, and sunflowers on a 320-acre farm. This year, the farmer wants to plant twice as many acres of to

svet-max [94.6K]

Set it up with variables; tomatoes=t, sunflowers=s, corn=c

320=t+s+c

This year;
t=2s
c=t+40

so, replace 2s for t; c=2s+40

now we can put those factors in for 320=t+c+s

320= 2s+2s+40+s
(SIMPLIFY)---> 320=5s+40----> 5s+280----> s=56

now incorporate the s value into the other equations.

c=2s+40----> c=56(2)=40 ----> c=152
t=2s -----> t=2(56) ----> t=112

TOMATOES= 112 acres
CORN= 152 acres
SUNFLOWERS= 56 acres

To check your work: 320=t+s+c ----> 320=112+156+56

5 0

2 years ago

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Westkost [7]

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Data;