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2 years ago

Given f(x) = 2x² + 4 and g(x) = 3x – 1 …… What is (fg)(x)? Show your work.

Mathematics

1 answer:

Lelu [443]2 years ago

6 0

Answer:

(fg)(x) = 6x³ - 2x² + 12x - 4

Step-by-step explanation:

( f g ) ( x ) = f ( x ) * g ( x )

f(x) = 2x² + 4

g(x) = 3x – 1

( f g ) ( x ) = (2x² + 4) (3x – 1)

= 6x³ - 2x² + 12x - 4

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Write an equation of the line that passes through (2, -5) and is parallel to the line 2y = 3x + 10

Zarrin [17]

First you get "y" by itself. To do so you divide 2 on both sides.

y = 3/2x + 5

To write an equation of a line that is PARALLEL to this equation, the slopes have to be the SAME. So the slope is 3/2.

You then use the equation:

y = mx + b

SInce you know "m" you plug it in.

y = 3/2x + b

Now you need to find b. To do so you plug in the point (2, -5) into this equation.

-5 = 3/2(2) + b

-5 = 3 + b

-8 = b

Finally you plug in b and you get your new equation.

y = 3/2x - 8

8 0

2 years ago

Geometry.

Harman [31]

Check the picture below.

8 0

2 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

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

Unit: Integration

4 0

2 years ago

100 points. Which is the graph of f (x) x - 1/ x2 - x 6

Dovator [93]

Answer: the answer is c

Step-by-step explanation:

5 0

1 year ago

Can someone help me please

goldenfox [79]

Answer:

$1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}$

Step-by-step explanation:

The applicable rule is ...

$x^{\frac{m}{n}}=\sqrt[n]{x^m}$

It works both ways, going from radicals to frational exponents and vice versa.

The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.

7 0

3 years ago