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3 years ago

If f(x)=2x+1 and g(x)=x2-7find(f+g)(x)

Mathematics

2 answers:

scoundrel [369]3 years ago

8 0

Answer:

$\boxed{\sf (f+g)(x) = {x}^{2} + 2x - 6}$

Given:

$\sf f(x) =2x + 1 \\ \sf g(x) = {x}^{2} - 7$

To find:

$\sf (f + g)(x) = f(x) + g(x)$

Step-by-step explanation:

$\sf \implies(f + g)x = f(x) + g(x) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (2x + 1) + ( {x}^{2} - 7) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x + 1 + {x}^{2} - 7 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} + 2x - 7 + 1 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {x}^{2} + 2x - 6$

docker41 [41]3 years ago

7 0

Answer:

x^2+2x -6

Step-by-step explanation:

f(x)=2x+1

g(x)=x^2-7

(f+g)(x)= 2x+1+x^2-7

Combine like terms

= x^2+2x -6

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3 years ago

Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)

zubka84 [21]

$\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } 
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This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

$\frac{ \sqrt{-49}}{7-2i-4-9i}$

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result:

$\frac{ \sqrt{-49} }{3-11i}$

Now, the $\sqrt{-49}$ must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to $\sqrt{49} \sqrt{-1}$
This means that we get the result 7i for the numerator.

$\frac{7i}{3-11i}$

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: $x^2 - y^2 =(x+y)(x-y)$
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

$\frac{7i (3+11i)}{(3-11i)(3+11i)}$

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how $(3-11i)(3+11i)= 3^2 -(11i)^2$ therefore, we can finally write:

$\frac{7i(3+11i)}{3^2 - (11i)^2 }$

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product $7i(3+11i) = 21i+77i^2$ .
You should know from your classes that i^2 = -1, thefore the numerator simplifies to $-77+21i$
You can do it as a curious thing, but simplifying yields the result:
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