All categories

2 years ago

Use the drop-down menus to identify the career that matches each description.

Mathematics

1 answer:

jolli1 [7]2 years ago

7 0

Answer:

D, B, C, and B

Step-by-step explanation:

I just did it

You might be interested in

A 5×5×5 wooden cube was painted and then sawed into 1×1×1 cubes. How many 1×1×1 cubes are there? HELPPPPP!!!!!!!!!!!!!!!!!!!!!!

Shalnov [3]

Answer:

125

Step-by-step explanation:

1 x 1 x 1 = 1

5 x 5 x 5 = 125

125 / 1 = 125

125 is your answer

6 0

3 years ago

Find the rate of markup if a book cost the retailer 17.50 and sells for $24.95. Round the the nearest tenth if needed

CaHeK987 [17]

Mark up = profit ÷ cost

Let M = mark up

Let p = profit

Let c = cost

M = p/c

Take it from here.

4 0

3 years ago

What is the inverse of the function f(x) = x +3?

MAXImum [283]

Answer:

f⁻¹(x) = x - 3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Functions
Function Notation
Inverse Functions

Step-by-step explanation:

Step 1: Define

Identify

f(x) = x + 3

Step 2: Find

Swap: x = y + 3
[Subtraction Property of Equality] Isolate y: x - 3 = y
Rewrite: f⁻¹(x) = x - 3

6 0

3 years ago

What is 0.08% written as a decimal?

otez555 [7]

Answer:

0.0008

Step-by-step explanation:

divide 0.08 by 100

answer is 0.0008

7 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

1 year ago