Answer:
We have the functions:
f(x) = IxI + 1
g(x) = 1/x^3.
Now, we know that the composite functions do not permute.
How we can prove this?
First, two composite functions are commutative if:
f(g(x)) = g(f(x))
Well, you could use brute force (just replace the values and see if the composite functions are commutative or not)
But i will use a more elegant way.
We can notice two things:
g(x) has a discontinuity at x = 0.
so:
f(g(x)) = I 1/x^3 I + 1
still has a discontinuty at x = 0, but:
g(f(x)) = 1/( IxI + 1)^3
here the denominator is IxI + 1, is never equal to zero.
So now we do not have a discontinuity.
Then the composite functions can not be commutative.
The unit of density [kgm^-3] displays the formula.
So density=mass/volume.
Convert 2160g into kilograms [2160/1000=2.160kg]
Convert volume of 240cm^3 into m^3
100cm=1m
100x100x100=1000000cm^3=1m^3
240/1000000=[2.4x10^-4]
Final calculation
2.160/2.4x10^-4=9000
Answer: 9000kgm^-3
Answer:
88.7364
Step-by-step explanation:
3.14 TIMES 28.26
<span>The <u>correct answers</u> are:
1, 3 and 4.
Explanation<span>:
We use synthetic division to divide a polynomial by a binomial of the form x-c, where c is a constant. x-c is a linear term, so 1 is true.
When we begin synthetic division, we place the value of c outside the box. Since c is a constant, and not a variable, then 2 is not true.
When we finish synthetic division, if there is a number other than 0 in the last term of the quotient (answer), this is a remainder. This makes 3 true.
When converting the answer from synthetic division to a polynomial, we use our numbers in the quotient as the coefficients of variables, starting with exponents that are 1 less than the original dividend; this makes 4 true.</span></span>
