3 is in the ones place. It's value is 3 ones or 3.
2 is in the tens place, so it's value is 2 tens or 20.
Equation: <span> y+7=-2/5(x-10)
y + 7 = -2/5 x + 4
y = -2/5 x - 3
2/5x + y = -3
Multiply the equation with 5,
2x + 5y = -15
In short, Your Answer would be: </span>2x + 5y = -15<span>
Hope this helps!</span>
Step-by-step explanation:
y = ax + b
I see already the result (y = x/10 × 1/6), but let's go in formally.
we have multiple function points to use to officially calculate a and b.
1/6 = a×10 + b
2/3 = a×40 + b
5/6 = a×50 + b
1 2/3 = a×100 + b
let's e.g subtract equation 1 from equation 3.
4/6 = a×40 + 0
a = 4/40 / 6 = 4 / 240 = 1/60
1/6 = 10/60 + b
1/6 = 1/6 + b
b = 0
so, the function is
y = x/60
x = 25
y = 25/60 = 5/12 cups
Answer:
x=15.5
Step-by-step explanation:
2x+1=32
2x=32-1
2x=31
x=15.5
The question regards composite functions. A composite function is a function composed of more than one function. Sorry for saying the word function so many times there, it's just what it is...
The phrase f(g(x)) means 'perform g on an input x, then perform f on the result'. You can then see that there are many options for f(x) and g(x) here, in fact an infinite number of one were to be ridiculous about it.
However a sensible choice might be g(x) = x^2, and f(x) = 2/x + 9. Checking:
g(x) = x^2
f(g(x)) = 2/(x^2) + 9
That is the first question dealt with. Next up is Q2. It is relatively simple to show that these functions are inverses. If you start with a value x, apply a function and then apply the function's inverse, you should return to the same starting value x. To take a common example, within a certain domain, sin^-1(sin(x)) = x.
f(g(x)) = (sqrt(3+x))^2 - 3 = 3 + x - 3 = x
g(f(x)) = sqrt(x^2 - 3 + 3) = sqrt(x^2) = x
A final note is that this is only true for a certain domain, that is x <= 0. This is because y = x^2 is a many-to-one function, so unrestricted it does not have an inverse. Take the example to illustrate this:
If x = -2, f(x) = (-2)^2 - 3 = 4 - 3 = 1
Then g(f(x)) =sqrt(1 + 3) = sqrt(4) = 2 (principal value).
However the question isn't testing knowledge of that.
I hope this helps you :)
