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gogolik [260]
2 years ago
6

A toy shop has stuffed Ana I’m als originally priced at $25 each. There is now a 20% discount on the stuffed animals. What is th

e sale price of each stuffed animal?
Mathematics
2 answers:
ICE Princess25 [194]2 years ago
8 0

Answer:

$20 for each stuffed animal

Step-by-step explanation:

[] The original price is $25, but it is 20% off.

100% - 20% = 80%

[] Now we can find 80% of $25.

-> First, 80% divided by 100 becomes 0.8, a decimal we can work with

25 * 0.8 = 20

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

tangare [24]2 years ago
4 0
$20 because you would also have to minus 20 from 100 and that’s 80
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Answer:

24

Step-by-step explanation

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1. Divide 140 by 6 and you get 23.3 repeating

2. When you multiply 23 times 6 you get 138. (not enough seats)

3. So, you would need to have at least 24 cars in order for everyone to go on the trip. (24 x 6 = 144 seats)

3 0
3 years ago
A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost
Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

S = 2πr² + 2πrh ( surface area of the lids + surface are of the curved surface)

Now the function that represents the cost to construct the can will be

C = 2πr²(0.027) + 2πrh(0.015)

C = 0.054πr² + 0.03πrh ---------(1)

Volume of the can = Volume of a cylinder = πr²h

16 = πr²h

h=\frac{16}{\pi r^{2}} -------(2)

Now we place the value of h in the equation (1) from equation (2)

C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})

C=0.054\pi r^{2}+0.03(\frac{16}{r})

C=0.054\pi r^{2}+(\frac{0.48}{r})

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

C'=0.108\pi r-(\frac{0.48}{r^{2} })

Now for C' = 0

0.108\pi r-(\frac{0.48}{r^{2} })=0

0.108\pi r=(\frac{0.48}{r^{2} })

r^{3}=\frac{0.48}{0.108\pi }

r³ = 1.415

r = 1.12 inch

and h = \frac{16}{\pi (1.12)^{2}}

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

C"=0.108+\frac{0.48}{r^{3}} which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

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3 years ago
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Answer:

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Step-by-step explanation:

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8 0
4 years ago
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Answer:

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LekaFEV [45]
X = x

to solve, subtract both sides by x:

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