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GaryK [48]
3 years ago
15

How to find the price of something before a discount?

Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0
Discount prices/the discount
E.g.the new price of the book is 75 which has a discount of 75%
The original price = 75/0.75=100
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What is the value,in degree, of y
Viktor [21]
Y= 122.

I did the math.
6 0
3 years ago
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25-14+3 my friends AND I DO NOT get it! Please help us! :)
irga5000 [103]
25-14=11 
11+3=14


25+3=28
28 divided by 2=14
8 0
3 years ago
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The area of the base of the regular quadrilateral pyramid is 36 cm2 and the area of a lateral face is 48 cm2. Find:
nikitadnepr [17]

Answer:

228 cm²

Step-by-step explanation:

The base of the pyramid is a regular quadrilateral (a square), so there are 4 congruent lateral faces.

The total surface area is therefore:

A = 36 cm² + 4 (48 cm²)

A = 228 cm²

5 0
4 years ago
How much tax is withheld from 53,620 if tax rate is 7%.?
Tems11 [23]

The withheld tax is 3753.4

To get this you do 53,620 x .07

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3 years ago
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Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-section
Angelina_Jolie [31]

Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

V = \int\limits^b_a {S(x)} \, dx          (1)

V = volume of the solid bounded by x = a and x = b

S(x) = cross section area of the solid, perpendicular to the x axis

From the figure we have that S is the area of a triangle that has base Z and height Y

Area of the triangle = S(x)=\frac{y(x)*z(x)}{2}          (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}          (3)

Equation of the line = y - y_{1} =m(x-x_{1} )          (4)

Replacing the points (1,0) and (0,1) in (3):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (4):

y-0=(-1)(x-1)

y(x) = -x + 1 (Line A-B)          (5)

We apply the equation of the point-slope line (plane xz):

Slope = m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}          (6)

Equation of the line = z - z_{1} =m(x-x_{1} )          (7)

Replacing the points (1,0) and (0,1) in (6):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (7):

z-0=(-1)(x-1)

z(x) = -x + 1 (Line A-C)        (8)

Replacing (5) and (8) in (2)

S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}          (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx

V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)  evaluated from x=0 to x=1

V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}

3 0
4 years ago
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