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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

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