Answer: −21.33333333333333
description: used BEDMAS
Brackets: (4 x 2/3)
division: 2/3 = 0.666666666666667
multiplication: 4x0.666666666666667= 2.666666666666667
Outside brackets: -8(2.666666666666667)
= −21.33333333333333
Answer:
P(x) = x³-(3i+1)x²+(3i-2)x+6i
Step-by-step explanation:
If -1, 2 and 3i are roots of the polynomial, then the following are factors of the polynomial in x
(x+1), x-2 and x-3i
To get the required polynomial, we will take the product of the factors as shown;
(x+1)(x-2)(x-3i)
(x+1)(x-2) = x²-2x+x-2
(x+1)(x-2) = x²-x-2
(x²-x-2)(x-3i) = (x²)(x)-x²(3i)-x(x)-x(-3i)-2x-2(-3i)
(x²-x-2)(x-3i) = x³-3x²i-x²+3xi-2x+6i
(x²-x-2)(x-3i) = x³-x²(3i+1)+x(3i-2)+6i
(x²-x-2)(x-3i) = x³-(3i+1)x²+(3i-2)x+6i
Hence the polynomial function in standard form with leading coefficient of 1 is x³-(3i+1)x²+(3i-2)x+6i
Answer:
21 ways
Step-by-step explanation:
number = 7 digit
5 digit no = 52115
to find out
How many different seven-digit numbers
solution
first we need to place the two missing 3s in the number 52115
we consider here two cases
case 1 the two 3's appear separated (like 532135 or 3521135)
case 2 the two 3's appear together (like 5332115 or 5211533)
Case 1 we can see that number type as _5_2_1_1_5_
place 3's placeholders show potential locations
( type a ) for 3's separated we will select 2 of 6 place and place 3 in every location so we do this 6C2 = (15) ways
and (type b): again use same step as _5_2_1_1_5_
here 3s together for criterion and we will select 1 of the 6 place and place both 3s here and there are 6 ways.
so that here will be 15+6=21 ways
If 3 and 3 are separate so 6C2 = 15 ways
If 3 and 3 are together so there = 6 ways
= 15 + 6 = 21 ways
Answer:
KJP is the answer, if I'm correct
