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3 years ago

Can someone help me with this problem

Mathematics

1 answer:

Maksim231197 [3]3 years ago

5 0

The area of a triangle is .5*b*h = .5x^2

.5x^2 = 27 so

x^2 = 54

x = sqrt(54) = 7.3

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What is the graph of the function f(x)=−⌊x⌋

kupik [55]

Answer:

Step-by-step explanation:

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2 years ago

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Brrunno [24]

Answer:

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Step-by-step explanation:

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3 years ago

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gogolik [260]

Answer:

Step-by-step explanation:

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3 years ago

Find the derivative of

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

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2 years ago

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ZanzabumX [31]

Answer: 2 7/10

Step-by-step explanation:

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2 years ago