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3 years ago

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suppose that f(x) = 4x+7 and g(x) =2x. what is a rule for g(f(x)? what is a rule for f(g(x))? suppose that f(x) = 4x+7 and g(x)

=2x. what is a rule for g(f(x)? what is a rule for f(g(x))?

1 answer:

Anastasy [175]3 years ago

4 0

The solution to the problem is as follows:

f o g = 8x + 7
<span>g o f = 2 ( 4x + 7 ) = 8x + 14
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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2 years ago

Read 2 more answers

Help me to answer now ineed this <br> Please...

Vera_Pavlovna [14]

ANSWER TO QUESTION 1

$\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}$

Let us change middle bar to division sign.

$\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}$

We now multiply with the reciprocal of the second fraction

$\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}$

We factor the first fraction using difference of two squares.

$\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}$

We cancel common factors.

$\frac{(y+2)}{(x-3)}\times \frac{1}{1}$

This simplifies to

$\frac{(y+2)}{(x-3)}$

ANSWER TO QUESTION 2

$\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}$

We change the middle bar to the division sign

$(1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})$

We collect LCM to obtain

$(\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}$

We expand and simplify to obtain,

$(\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}$

$(\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}$

We now multiply with the reciprocal,

$(\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}$

We cancel out common factors to obtain;

$(\frac{1}{x})\times \frac{(x+2)(x+3)}{1}$

This simplifies to;

$\frac{(x+2)(x+3)}{x}$

ANSWER TO QUESTION 3

$\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}$

We rewrite the above expression to obtain;

$\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}$

We now multiply by the reciprocal,

$\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}$

We multiply out to get,

$\frac{(a-b)^2}{(a+b)^2}$

ANSWER T0 QUESTION 4

To solve the equation,

$\frac{m}{m+1} +\frac{5}{m-1} =1$

We multiply through by the LCM of $(m+1)(m-1)$

$(m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1$

This gives us,

$(m-1) \times m + (m+1) \times 5}=(m+1)(m-1)$

$m^2-m+ 5m+5=m^2-1$

This simplifies to;

$4m-5=-1$

$4m=-1-5$

$4m=-6$

$\Rightarrow m=-\frac{6}{4}$

$\Rightarrow m=-\frac{3}{2}$

ANSWER TO QUESTION 5

$\frac{3}{5x}+ \frac{7}{2x}=1$

We multiply through with the LCM of $10x$

$10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1$

We simplify to get,

$2 \times 3+5 \times 7=10x$

$6+35=10x$

$41=10x$

$x=\frac{41}{10}$

$x=4\frac{1}{10}$

Method 1: Simplifying the expression as it is.

$\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}$

We simplify further to get,

$\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}$

$\frac{\frac{19}{20}}{\frac{37}{40}}$

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator

$\frac{\frac{19}{1}}{\frac{37}{2}}$

Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.

That is;

$\frac{19\times 2}{1\times 37}$

This simplifies to

$\frac{38}{37}$

Method 2: Changing the middle bar to a normal division sign.

$(\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$(\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})$

We simplify further to get,

$(\frac{15+ 4}{20})\div (\frac{(25+12)}{40})$

$\frac{19}{20}\div \frac{(37)}{40}$

We now multiply by the reciprocal,

$\frac{19}{20}\times \frac{40}{37}$

$\frac{19}{1}\times \frac{2}{37}$

$\frac{38}{37}$

5 0

3 years ago