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Doss [256]
3 years ago
8

A farmer plants twice as many acres of wheat as corn on his land. If he plants no more than 36 acres, what is the greatest numbe

r of acres that can be used for corn? 12 acres 18 acres 24 acres 3 acres
Mathematics
1 answer:
Leona [35]3 years ago
6 0

Answer: 12 acres could be planted (at most)

Step-by-step explanation: Since there will be twice as many acres of wheat in  comparison to the corn, out of the total 36 acres at least 24 of those will need to be wheat. That leaves 12 acres left for the corn.

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Pls send me step by step pic​
Thepotemich [5.8K]

To solve, simply do this:

(\frac{1}{2}^3)-(\frac{3}{4} )^2\\\\\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} \\\\\frac{1}{8} -(\frac{3}{4})^2\\\\\frac{1}{8}-\frac{9}{16}\\\\=\frac{-7}{16}

Then you'll get the answer, -7/16

7 0
3 years ago
NEED HELP ASAP PLEASE
AnnZ [28]

Answer:

8.5%

Step-by-step explanation:

I=PRT

85= 2000*r*0.5

85=1000r

0.085=r

8.5%

6 0
3 years ago
Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
What is the solution to y=-x+3 and y=4x-2
DIA [1.3K]

Answer: x=1

              y=2

Step-by-step explanation:

y=-x+3

y=4x-2

-----------

    y=y

-x+3=4x-2

3+2=4x+x

5=5x

x=1

y=-1+3

y=2

3 0
3 years ago
What is the solution of this system of linear equations?
shutvik [7]

Answer:

funny I have this question and I thought it was D

6 0
2 years ago
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