Answer:
Linear functions are equations where the greatest power of the variable (typically x) is 1;
They are graphed as straight lines;
So, any function that has the following format:
f(x) = ax + b;
Nonlinear functions would be any equation that has a variable to a power greater than 1;
This could be a quadratic equation or an exponential equation
T=-16... I hope this helps love! :)
Answer:
146.41
Step-by-step explanation:
third order determinant = determinant of 3×3 matrix A
given ∣A∣=11
det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)
=det(adjacent of A)
{det (cofactor matrix of A)} ^2 = {det (adjacent of A)}
^2
(Using for an n×n det (cofactor matrix of A)=det (A)^n−1
)
we get
det (cofactor matrix of A)^2 = {det(A) ^3−1
}^2
=(11)^2×2 = 11^4
=146.41
4. Let the numbers be x - 2, x, and x + 2, where x is an odd number.
2(x² - 4) - 4x = (x + 2)² + 21
2x² - 8 - 4x = x² + 4x + 4 + 21
2x² - 4x - 8 = x² + 4x + 25
x² - 8x - 33 = 0
(x - 11)(x + 3) = 0
x = 11, or -3
When x = 11, x - 2 = 9, x + 2 = 13
When x = -3, x - 2 = -5, x + 2 = -1
Explanation: You are on the right track. However, rather than having three unknown variables, try to reduce your working out to one unknown variable. Since you know they are consecutive odd numbers, you can simply let x be the middle term and the other two be + and - 2, provided x is an odd number.
That will reduce your variable issues, and helps as the first and third provide a difference of two squares, and this works out very nicely.
Q5: is essentially the same process. Let your variables be something in the form of one unknown variable, and you should be okay from there. Let me know if you're stuck.
8 times 3 = 24 +4= 28
4+3=7 times 8 =56
