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1 year ago

Can someone please help me with this for brainliest!!!!!

Mathematics

1 answer:

sergiy2304 [10]1 year ago

4 0

Answer:

Step-by-step explanation:

(f ⁰ g)(x) = 3(x² + 4) + 5

(f ⁰ g)(4) = 3(4² + 4) + 5 = 3*20 + 5 = 65

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Help me ahh eeeeee Image is problem/question

Sphinxa [80]

Answer:

(4,1)

Step-by-step explanation:

point b is at (-1, 4) and the rule for 90 degree clockwise rotations is (y, -x) so it's (4, 1)

6 0

2 years ago

Read 2 more answers

PLEASE ANSWER THESE I RLLY NEED HELP I DONT UNDERSTAND I WILL MARK YOU AS BRAINLIEST IF YOU ANSWRR CORRECTLY PLS AND TY

faust18 [17]

Answer:

Step-by-step explanation:

5. You are asked to write an equation of the line in slope-intercept form, so you need to determine the slope of the line and the y-intercept.

You're lucky. One of the points, (0,1), has an x-coordinate of 0, so you know that the y-intercept is 1.

Use the coordinates of the points to determine the slope of the line.

slope = (difference in y-coordinates)/(difference in x-coordinates) = (10-1)/(3-0) = 9/3 = 3

The slope-intercept equation of the line is y = 3x+1

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7. When x = 0, function A = 0 and function B = 3, so function B has a greater initial value.

7 0

2 years ago

Expand or factor each of the following expressions to determine which expressions are equivalent.

bearhunter [10]

(3x-4y)^2=9x^2-24x+16y^2
(3x-2)(9x^2+6x+4)=27x^3-8

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2 years ago

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Given that f(–2.4) = -1 and f(-1.9) = -8, approximate f'(-2.4). f'(-2.4)

lora16 [44]

Answer:

f'(-2.4) ≈ -14

General Formulas and Concepts:
Algebra I

Coordinate Planes

Coordinates (x, y)

Slope Formula: $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

Functions

Function Notation

Calculus

Differentiation

Derivatives
Derivative Notation

Step-by-step explanation:

*Note:

The definition of a derivative is the slope of the tangent line.

Step 1: Define

Identify.

f(-2.4) = -1

f(-1.9) = -8

Step 2: Differentiate

Simply plug in the 2 coordinates into the slope formula to find slope m.

[Derivative] Set up [Slope Formula]: $\displaystyle f'(-2.4) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}$
Substitute in coordinates: $\displaystyle f'(-2.4) \approx \frac{-8 - -1}{-1.9 - -2.4}$
Evaluate: $\displaystyle f'(-2.4) \approx -14$

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Learn more about derivatives: brainly.com/question/17830594

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

Unit: Differentiation

6 0

1 year ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

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

Unit: Differentiation

7 0

1 year ago