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BaLLatris [955]

3 years ago

9

Kira's store collects 5% sales tax on every item sold. If she collected $22 in sales tax, what was the cost of the items her sto

re sold?

1 answer:

Volgvan3 years ago

4 0

Answer:

$\$440$

Step-by-step explanation:

Let

x -----> the cost of the items Kira's store sold

Remember that

$5\%=5/100=0.05$ ----> sales tax in decimal number

we know that

The cost of the items that Kira's store sold multiplied by the sales tax in decimal number must be equal to $22

so

$0.05x=22$

Solve for x

Divide by 0.05 both sides

$x=\$440$

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2/9 multiplied by 9/5

r-ruslan [8.4K]

Answer:

2/5

Step-by-step explanation:

simplify by cross multiplying

7 0

2 years ago

Read 2 more answers

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

What is the value of v

const2013 [10]

Answer:

v=9

Step-by-step explanation:

12v-44+6v+23+15v-9+v+30+6v=360

40v=360

v=9

6 0

3 years ago

What radius of a circle is required to inscribe a regular hexagon with an area of 210.44 cm2 and an apothem of 7.794 cm

alekssr [168]

We know that

[area of a regular hexagon]=6*[area of one <span>equilateral triangle]
</span>210.44=6*[area of one equilateral triangle]
[area of one equilateral triangle]=210.44/6-----> 35.07 cm²

[area of one equilateral triangle]=b*h/2
h=7.794 cm
b=2*area/h------> b=2*35.07/7.794------>b= 9 cm

the length side of a regular hexagon is 9 cm
<span>applying the Pythagorean theorem
</span>r²=h²+(b/2)²------>r²=7.794²+(4.5)²------> r²=81--------> r=9 cm

<span>this last step was not necessary because the radius is equal to the hexagon side------> (remember the equilateral triangles)
</span>
the answer is
the radius is 9 cm

8 0

3 years ago

The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization’s staff of

pishuonlain [190]

Answer:

Null Hypothesis, $H_0$ : p $\geq$ 45%

Alternate Hypothesis, $H_A$ : p < 45%

Step-by-step explanation:

We are given that the proportion of women in similar sales positions across the country in 2004 is less than 45%.

The collected random sample of size 50 showed that only 18 were women.

<u><em>Let p = proportion of women in similar sales positions across the country in 2004</em></u>

So, Null Hypothesis, $H_0$ : p $\geq$ 45%

Alternate Hypothesis, $H_A$ : p < 45%

Here, <u><em>null hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is more than or equal to 45%.

On the other hand, <u><em>alternate hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is less than 45%.

The test statistics that would be used here is One-sample z proportion test statistics, i.e;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hta p(1-\hat p)}{n} } }$ ~ N(0,1)

Hence, the above hypothesis is appropriate for the given situation.

4 0

3 years ago