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lana [24]
3 years ago
9

Help fast not sure on this question. Jeff can weed the garden twice as fast as his sister Julia. Together they can weed the gard

en in 3 hours. How long would it take each of them working alone?
A) 4.5 hours; Julia 9 hours
B) 7.5 hours; Julia 15 hours
C) 4.5 hours; Julia 2.25 hours
D) 0.5 hours; Julia 1 hours
Mathematics
1 answer:
aksik [14]3 years ago
8 0

Answer:

c its the only one that makes sense

Step-by-step explanation:

jeff dose more weeds then julia and the only one were this is tru

if im right i get brainliest

You might be interested in
Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.
elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}

Consider a vector field F=x\^i+y\^j+z\^k

Then the divergence of the vector F is,

div F = \nabla.F = (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)

and the curl of the vector F is,

curl F = \nabla\times F = \^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})

<h3>Calculation:</h3>

The given vector fields are:

F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k

1) Verifying the identity: \nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

Consider L.H.S

⇒ \nabla.(aF1+bF2)

⇒ \nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))

⇒ \nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the dot product between these two vectors,

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(1)

Consider R.H.S

⇒ a\nabla.F1+b\nabla.F2

So,

\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)

⇒ \nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}

\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)

⇒ \nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}

Then,

a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(2)

From (1) and (2),

\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

2) Verifying the identity: \nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Consider L.H.S

⇒ \nabla\times(aF1+bF2)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the cross product,

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y}) ...(3)

Consider R.H.S,

⇒ a\nabla\times F1+b\nabla\times F2

So,

a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))

⇒ \^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})

a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))

⇒ \^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})

Then,

a\nabla\times F1+b\nabla\times F2 =

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})

...(4)

Thus, from (3) and (4),

\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

8 0
1 year ago
Radical 8 plus radical 18
maxonik [38]

Answer:

5√2

Step-by-step explanation:

√8 + √18

We first have to find what is the largest perfect square that goes into √8:

4 is the largest, so therefore → √8 gives you 2√2:

Work: √4 * √2 → 2 * √2 → 2√2

Now we have to find what is the largest perfect square that goes into √18:

9 is the largest, so therefore → √18 gives you 3√2:

Work: √9 * √2 → 3 * √2 → 3√2

Because 2√2 and 3√2 have the same "base" of √2, they can be added together:

2√2 + 3√2 = 5√2 (The "bases" are to be left alone!)

7 0
3 years ago
You accept a new job with a starting salary of $47,000. You receive a 4% raise at the start of your second year, a 5.6% raise at
Sunny_sXe [5.5K]

Answer:

1yr  47,000, 2nd yr  48,880, 3rd yr   51,617.28, 4th yr  57,346.7981 or 57,346.80

Step-by-step explanation:

1yr        47,000

2nd yr  47,000 x 4% = 48,880

3rd yr   48,880 x 5.6% = 51,617.28

4th yr   51,617.28 x 11.1% = 57,346.7981 or 57,346.80

4 0
2 years ago
What equation is graphed in this figure?
topjm [15]

Answer:

The third equation: y+2=-3(x-1)

Step-by-step explanation:

The two points on the line are (-1,4) and (1,-2).

Slope of the line passing through two points (x_{1},y_{1}) and (x_{2},y_{2}) is given as:

m=(y_{2} -y_{1})/ (x_{2} -x_{1})

Here, (x_{1},y_{1}) and (x_{2},y_{2}) are (-1,4) and (1,-2).

Therefore, slope is equal to, m=(-2-4 )/ (1-(-1)

                                               m=-6/2

                                               m=-3  

Now, equation of a straight line with slope m and points (x_{1},y_{1}) and (x_{2},y_{2}) is given as:

y-y_{1}=m(x-x_{1})\\y-y_{2}=m(x-x_{2})

Now, if we use the 2nd form, then x_{2}=1,y_{2}=-2.

So, the equation is given as :

y-(-2)=-3(x-1)\\y+2=-3(x-1)

                                                 

8 0
3 years ago
Determine whether the given graph is a function or not.
BaLLatris [955]

Answer:

Function

Step-by-step explanation:

Yes, the graph is a function. For each x value, there is a unique y value that corresponds with it.

A graph can also be tested as a function by using the popular "vertical-line test".

5 0
1 year ago
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