Answer:
See explanations below
Step-by-step explanation:
Example of how to find the determinant of a 2×2 matrices is as shown;
![\left[\begin{array}{ccc}a&b\\c&d&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%26%5Cend%7Barray%7D%5Cright%5D)
= (a*d)-(b*c)
= ad - bc
Applying this to solve the given questions
1) (3*4) - (-3*-1)
= 12 - (3)
= 9
2) (-12*5)-(-1*12)
= -60 - (-12)
= -60+12
= -48
3) (0*4)-(-1*17)
= 0-(-17)
= 0+17
= 17
4) (8*-1)-(-4*-1)
= -8-(4)
= -8-4
= -12
5) (2*-1)-(2*-1)
= -2-(-2)
= -2+2
= 0
6) 1(1) - 0(0)
= 1 - 0
= 1
Answer: 8y^6x^6
Step-by-step explanation:
Volume = l*w*h so multiply all the variable by <u>multiplying the numbers and adding the exponents</u> so
2y^2*4x^4y=8y^3x^4
then 8y^3x^4*x^2y^3=8y^6x^6
Answer:x=10
Step-by-step explanation:x x2
Take 3 4 5 3 4 5 6 5 4 3 2 3 4 5 6 4 8 4 3 2 and add them all together
=83
Now you divide 83 by the number of terms in the sample set, which is 20.
83/20 = 4.15
For rounding to the nearest tenth, look at the hundredths space, which is 5. Since it is 5 or higher, round the tenths up one.
final answer= 4.2
Answer:
6, 2, 2/3, 2/9, 2/27, 2/81
Step-by-step explanation:
The nth term of a geometric progression is expressed as;
Tn = ar^n-1
a is the first term
n is the number of terms
r is the common ratio
Given
a = 6
r = 1/3
when n = 1
T1 = 6(1/3)^1-1
T1 = 6(1/3)^0
T1 = 6
when n = 2
T2= 6(1/3)^2-1
T2= 6(1/3)^1
T2 = 2
when n = 3
T3 = 6(1/3)^3-1
T3= 6(1/3)^2
T3= 6 * 1/9
T3 = 2/3
when n = 4
T4 = 6(1/3)^4-1
T4= 6(1/3)^3
T4= 6 * 1/27
T4 = 2/9
when n = 5
T5 = 6(1/3)^5-1
T5= 6(1/3)^4
T5= 6 * 1/81
T5 = 2/27
when n = 6
T6 = 6(1/3)^6-1
T6= 6(1/3)^5
T6= 6 * 1/243
T6 = 2/81
Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81
