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2 years ago

Find (f+g)(x)f(x) = 4x−4 and g(x)=2x2−3x

Mathematics

2 answers:

skad [1K]2 years ago

8 0

Answer:

(f + g)(x) = 2x² + x - 4

Step-by-step explanation:

f(x) = 4x - 4

g(x) = 2x² - 3x

(f + g)(x) is the sum of functions f(x) and g(x).

(f + g)(x) = f(x) + g(x)

(f + g)(x) = 4x - 4 + 2x² - 3x

Combine like terms.

(f + g)(x) = 2x² + x - 4

marusya05 [52]2 years ago

5 0

The resulting expression for the sum of the function is 2x^2 - x - 4

<h3>Sum of function</h3>

Given the following function as shown below;

f(x) = 4x - 4

g(x) = 2x^2 - 3x

We are to determine the composite function (f+g)(x)

Substitute

(f+g)(x) = f(x) + g(x)

(f+g)(x) =. 4x-4 + 2x^2 -3x

(f+g)(x) = 2x^2 - x - 4

Hence the resulting expression for the sum of the function is 2x^2 - x - 4

Learn more on sum of function here: brainly.com/question/11602229

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nadya68 [22]

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

$\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}$

in this problem we have

$\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})$

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

$g^{2}=h^{2}+i^{2}-2*h*i*cos\ G$

In this problem we have

g=HI

G=angle Beta

substitute

$HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta$

$HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}$

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the answer Part 2) is

Law of Cosines

3 0

3 years ago

Read 2 more answers

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salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$