3/10+2 1/10+5 1/10 I need help

Solve the inequality |x| < 5. {x|-5 < x < 5} {x|x < -5 ∪ x > 5} {x|x -5 ∪ x < 5}

sesenic [268]

ANSWER

The correct answer is A

{ $x|-5$ }

EXPLANATION

$|x|$

This implies that

$x$

We solve the compound inequality to obtain,

$x-5$

Remember that, dividing through by negative 1 reverses the inequality sign

The solution set is

{ $x|-5$ }

8 0

3 years ago

For her tutoring services, Thuy charged Demi $10 per hour and $8 for books and supplies. Demi paid a total of $48.00. Which equa

Fudgin [204]

To find the final amount, you would have to multiply her rate per hour (10) by how many hours she tutored (h). Then, you need to add her flat rate of 8.

10h + 8 = 48, where h is how many hours she tutored.

7 0

2 years ago

Evaluate 2d+32 when d=8.

Sloan [31]

Given $2d+32$

$d=8$

Put the value of d in the given equation

$2d+32$

$2(8)+32$

$16+32$ $= 48$

What we do in Evaluation in math?

The verb "evaluate" is used in mathematics to mean to discover or determine the worth of something. Finding the precise value of an algebraic expression or numerical computation is relevant here as well as financial mathematics.

In mathematics, the terms "evaluate" and "solve" are often used interchangeably. Consider the following basic multiplication problem: "Evaluate the cost of six tomatoes when the tomatoes are purchased at $2 each." The student would multiply 6 by 2 to get the result of $12. The term evaluate might also come before a list of inquiries, as a list of straightforward algebraic equations.

Learn more about Evaluate brainly.com/question/23574746

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8 0

1 year ago

If the gradient of the tangent to

Marizza181 [45]

Answer:

Point A(9, 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

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$