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ddd [48]
2 years ago
6

Kelly wants to buy a dress that costs $40.00. She has a coupon for 25% off, and

Mathematics
1 answer:
Sophie [7]2 years ago
3 0

A percentage is a way to describe a part of a whole. The amount that Kelly will pay for the dress is $33.

<h3>What are Percentages?</h3>

A percentage is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25 which is equal to 25%.

To convert a fraction to a percentage, convert the fraction to decimal form and then multiply by 100 with the '%' symbol.

Given Kelly wants to buy a dress that costs $40.00. She has a coupon for 25% off, and there is a 10% sales tax. Therefore, the amount she will pay is,

25% of $40.00 is,

0.25 × $40 = $10

After using the coupon, the sale price is,

$40 - $10 = $30

The 10% tax on the sale price is,

0.10 × $30 = $3

The final price is $

$30 + $3 = $33

Hence, the amount that Kelly will pay for the dress is $33.

Learn more about Percentages:

brainly.com/question/6972121

#SPJ1

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Answer:

Step-by-step explanation:

1km = 1000m

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5e. 17.658km = 17658m = 17.7m

To round to the nearest 100m just divide by 100 and then round it off. like 57.68=57.7m because 8 is more than 5 so 6 becomes 7 to one decimal.

Rule is applied for all the other questions too.

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3 years ago
Simplify the fraction completely.
k0ka [10]

Answer:

5

Step-by-step explanation:

40% means 40 parts out of 100. If we have made the numerator of the fraction 2, we have divided 40 by 20. To make this fraction equal to 40/100, we have to divide 100 by 20 as well, giving us 5.

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2 years ago
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For number 1, the median is 8
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What is the equation of this line?
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For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

3 0
3 years ago
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