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dem82 [27]
2 years ago
5

If f(x) = x + 8 and g(x) = –4x – 3, find (f + g)(x).

Mathematics
2 answers:
Soloha48 [4]2 years ago
6 0

Answer:

\huge\boxed{\sf (f + g)(x) = -3x + 5}

Step-by-step explanation:

<h3><u>Given functions:</u></h3>
  • f(x) = x + 8
  • g(x) = -4x - 3
<h3><u>Solution:</u></h3>

Add both functions

(f + g)(x) = x + 8 + (-4x - 3)

(f + g)(x) = x + 8 - 4x - 3

(f + g)(x) = x - 4x + 8 - 3

(f + g)(x) = -3x + 5

\rule[225]{225}{2}

Alecsey [184]2 years ago
5 0

Answer:

B

Step-by-step explanation:

(f + g)(x)

= f(x) + g(x)

= x + 8 - 4x - 3 ← collect like terms

= - 3x + 5

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den301095 [7]

Answer:

see explanation

Step-by-step explanation:

Assuming the fractions are being multiplied

Factorise the denominators of both fractions

x² - 3x - 10 = (x - 5)(x + 2)

x² + x - 12 = (x + 4)(x - 3)

The product can now be expressed as

\frac{x-5}{(x-5)(x+2)} × \frac{x+2}{(x+4)(x-3)}

Cancel (x - 5) and (x + 2) on the numerators/ denominators, leaving

\frac{1}{(x+4)(x-3)} = \frac{1}{x^2+x-12}

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Step-by-step explanation:

Search for questions & chapters

Class 11

>>Maths

>>Trigonometric Functions

>>Trigonometric Equations

>>Solve for x: sin x + sin 2x...

Question

Bookmark

Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.

Medium

Solution

verified

Verified by Toppr

We write the given equation as

(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x

or 2sin2xcosx+sin2x=2cos2xcosx+cos2x

or sin2x(2cosx+1)=cos2x(2cosx+1)

or (sin2x−cos2x)(2cosx+1)=0

∴sin2x−cos2x=0 or 2cosx+1=0

If sin2x−cos2x=0, then tan2x=1,

Hence 2x=nπ+π/4

or x=(4n+1)

8

π

.(1)

If 2cosx+1=0, then cosx=−1/2 (2)

∴x=2nπ±

3

2π

or x=

3

6n±2

π

We seek values of x in the interval 0≤x≤2π

In this interval (1) gives

x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)

and (2) gives x=2π/3,4π/3. (for n=0,1)

Thus we get the answer

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

4 0
2 years ago
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