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3 years ago

What separates the portion of a number that is greater than 1 from the portion of the number that less than 1

Mathematics

1 answer:

konstantin123 [22]3 years ago

7 0

Answer:

1/2 -2/1

Step-by-step explanation:

122

8mar12

212323

123345wqesfef

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3 years ago

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Simplify.<br> (WILL GIVE BRAINLIEST)

Novosadov [1.4K]

Answer:

$\frac{\sqrt{x} }{2x}$

Step-by-step explanation:

Rationalize the denominator first (keep in mind we are NOT solving this because there is no equals sign here. We are merely simplifying.)

$\frac{\sqrt{5x} }{x\sqrt{20} } *\frac{\sqrt{20} }{\sqrt{20} } =\frac{\sqrt{100x} }{20x}$

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3 years ago

11.What is the solution of the inequality below?<br> |2x + 7| > 9

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3 years ago

For the following pair of functions, find (f+g)(x) and (f-g)(x).

ch4aika [34]

Given:

The functions are

$f(x)=4x^2+7x-5$

$g(x)=-9x^2+4x-13$

To find:

The functions $(f+g)(x)$ and $(f-g)(x)$ .

Solution:

We know that,

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

$(f+g)(x)=4x^2+7x-5-9x^2+4x-13$

$(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)$

$(f+g)(x)=-5x^2+11x-18$

And,

$(f-g)(x)=f(x)-g(x)$

$(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)$

$(f+g)(x)=4x^2+7x-5+9x^2-4x+13$

$(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)$

$(f-g)(x)=13x^2+3x+8$

Therefore, the required functions are $(f+g)(x)=-5x^2+11x-18$

and $(f-g)(x)=13x^2+3x+8$ .

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3 years ago