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statuscvo [17]
3 years ago
7

The width of a rectangle is 5 centimeters less than triple the length. Let m be length of the rectangle. Write a variable expres

sion for the width.
Mathematics
1 answer:
Olenka [21]3 years ago
4 0

length: m

less means "subtract", than means "reverse the order", triple means "times 3"

width: 3m - 5

Answer: 3m - 5


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Determine the domain of the function (fog)(x) where f(x)=3x-1/x-4 and g(x)=x+1/x
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Answer: (-∞,-1) ∪ (0,+∞)

Step-by-step explanation: The representation fog(x) is a representation of composite function, meaning one depends on the other.

In this case, fog(x) means:

fog(x) = f(g(x))

fog(x) = 3(x+\frac{1}{x} )-\frac{1}{x+\frac{1}{x} } -4

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

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

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

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

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

This is the function fog(x).

The domain of a function is all the values the independent variable can assume.

For fog(x), denominator can be zero, so:

x(x^{2}+x) \neq 0

If x = 0, the function doesn't exist.

x^{2}+x \neq0

x(x+1) \neq0

x+1\neq0

x\neq-1

<u>Therefore, the domain of this function is: </u><u>-∞ < -1 or x > 0</u>

5 0
3 years ago
Tan^2x+sec^2x=1 for all values of x. <br> true or false?
amid [387]

This is not true.

\tan^2x+\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\cos^2x+1}{\cos^2x}=2\sec^2x-1

2\sec^2x-1=1\implies\sec^2x=1\implies\sec x=\pm1\implies x=n\pi

where is n is any integer. So suppose we pick some value of x other than these, say x=\dfrac\pi4. Then

\tan^2\dfrac\pi4+\sec^2\dfrac\pi4=1+2=3\neq1

5 0
3 years ago
Read 2 more answers
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