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3 years ago

37 1/2 % as a decimal

Mathematics

2 answers:

Marysya12 [62]3 years ago

7 0

Answer: 37.5

1/2 in decimal from is 0.50 or 0.5

just olya [345]3 years ago

5 0

The answer is 37.5 because 37 is a whole number then add your decimal behind that. You get 37. then you add a half which equals.5 so put the 5 after the 37. and you get 37.5

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Learn with an example

mafiozo [28]

Answer:

-8

Step-by-step explanation:

In order to find the answer to this question you need to use the slope formula, substitute, subtract, and simplify if needed.

$(9,5),(4,1) = (x^1,y^1), (x^2,y^2)$

$m = \frac{y^2-y^1}{x^2-x^1} = \frac{9-1}{4-5}$

$9-1=8$

$4-5=-1$

$= -8$

Hope this helps.

7 0

2 years ago

School QUICK CARE

grandymaker [24]

Answer:

Step-by-step explanation:

Total pencils = 48

Total erasers = 72

If we put 1 pencil and 1.5 eraser in each pouch then total pouches = 48 pouches

3 0

3 years ago

A. Use composition to prove whether or not the functions are inverses of each other.

kogti [31]

A. In a composition of two functions the first function is evaluated, and then the second function is evaluated on the result of the first function. In other word, you are going to evaluate the second function in the first function.

Remember that you can evaluate function at any number just replacing the variable in the function with the number. For example, let's evaluate our function $f(x)$ at $x=1$ :

$f(x)=\frac{1}{x-3}$

$f(1)=\frac{1}{1-3}$

$f(1)=\frac{1}{-2}$

Similarly, to find the composition of $f(x)$ and $g(x)$ , we are going to evaluate $f(x)$ at $g(x)$ . In other words, we are going to replace $x$ in $f(x)$ with $\frac{3x+1}{x}$ :

$f(x)=\frac{1}{x-3}$

$f(g(x) = f(\frac{3x+1}{x} ) = \frac{1}{\frac{3x+1}{x} -3}$

Remember that two functions are inverse if after simplifying their composition, we end up with just $x$ . Let's simplify and see what happens.

$f(g(x)=\frac{1}{\frac{3x+1}{x} -3}$

$f(g(x)=\frac{1}{\frac{3x+1-3x}{x} }$

$f(g(x)=\frac{1}{\frac{1}{x} }$

$f(g(x)=x$

Now let's do the same for $g(f(x))$ :

$g(\frac{1}{x-3} )=\frac{3(\frac{1}{x-3})+1}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{3}{x-3}+1}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{3+x-3}{x-3}}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{x}{x-3}}{x}$

$g(\frac{1}{x-3} )=\frac{x}{x(x-3)}$

$g(f(x))=\frac{x}{x(x-3)}$

We can conclude that $g(x)$ is the inverse function of $f(x)$ , but $f(x)$ is not the inverse function of $g(x)$ .

B. The domain of a function is the set of all the possible values the independent variable can have. In other words, the domain are all the possible x-values of function.

Now, interval notation is a way to represent and interval using an ordered pair of numbers called the end points; we use brackets [ ] to indicate that the end points are included in the interval and parenthesis ( ) to indicate that they are excluded.

Notice that when $x=0$ , $g(x)=\frac{3(0)+1}{0} =\frac{0}{0}$ , so when $x=0$ , $g(x)$ is not defined; therefore we have to exclude zero from the domain of $f(g(x))$ .

We can conclude that the domain of the composite function $f(g(x))$ in interval notation is (-∞,0)U(0,∞)

Now let's do the same for $g(f(x))$ .

Notice that the composition is not defined when its denominator equals zero, so we are going to set its denominator equal to zero to find the values we should exclude from its domain:

$x(x-3)=0$