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2 years ago

A pair of boots is on sale for 20% off. The regular price of the boots is $13. What is the sale price of the boots?

Mathematics

1 answer:

Nady [450]2 years ago

6 0

Answer: $10.40

Step-by-step explanation:

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what is the volume of a rectangular prism that has a length of 8.5 centimeters (cm), a width of 3.2 centimeters, and a height of

Yuki888 [10]

Answer:

The volume of the rectangular prism is $163.2cm^3$

Step-by-step explanation:

A rectangular prism is in the shape of cuboid which has three dimensions i.e Length, width and height.

Given: Length =8.5 cm width= 3.2 cm and Height= 6 cm

The volume of a Cuboid = $Length*Width*Height$

Applying the given dimensions in the formula:

The volume of a Cuboid= $8.5cm*3.2cm*6cm$

= $163.2cm^3$

The volume of the rectangular prism is $163.2cm^3$

3 0

3 years ago

4/65 with remeinders

zavuch27 [327]

I think it’s 16 remainder 1

4 0

3 years ago

Use the function rule y=5x+1. What is y when x=3

Dmitry_Shevchenko [17]

Answer:

y = 16

(3, 16)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

y = 5x + 1

x = 3

Step 2: Evaluate

Substitute in x: y = 5(3) + 1
Multiply: y = 15 + 1
Add: y = 16

7 0

3 years ago

Read 2 more answers

Kyle and his brother have a marble set. Of the marbles 12 are blue this represents 50/100of ALL the marbles. What decimal is equ

Alex73 [517]

To find this you would do 50/100 which equals .5

3 0

3 years ago

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

4 0

3 years ago