(1 - 2x)⁴
(1 - 2x)(1 - 2x)(1 - 2x)(1 - 2x)
[1(1 - 2x) - 2x(1 - 2x)][1(1 - 2x) - 2x(1 - 2x)]
[1(1) - 1(2x) - 2x(1) - 2x(-2x)][1(1) - 1(2x) - 2x(1) - 2x(-2x)]
(1 - 2x - 2x + 4x²)(1 - 2x - 2x + 4x²)
(1 - 4x + 4x²)(1 - 4x + 4x²)
1(1 - 4x + 4x²) - 4x(1 - 4x + 4x²) + 4x²(1 - 4x + 4x²)
1(1) - 1(4x) + 1(4x²) - 4x(1) - 4x(-4x) - 4x(4x²) + 4x²(1) - 4x²(4x) + 4x²(4x²)
1 - 4x + 4x² - 4x + 16x² - 16x³ + 4x² - 16x³ + 16x⁴
1 - 4x - 4x + 4x² + 16x² + 4x² - 16x³ - 16x³ + 16x⁴
1 - 8x + 24x² - 32x³ + 16x⁴
Answer:
should be 5
Step-by-step explanation:
I assume the first equation is supposed to be
and not
As an augmented matrix, this system is given by
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C4%26-2%264%2612%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/2:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C2%26-1%262%266%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%265%265%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/5:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 3) to row 1, and add 3(row 3) to row 2:
![\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%260%2611%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -3(row 2) to row 1:
![\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D-1%260%260%26-1%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%26-1%260%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multipy through row 2 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%261%260%26-2%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
The solution to the system is then
Answer:
3 peanut cookies
Step-by-step explanation:
Given that :
Plate 1:
Number of chocolate chip = 4
Number of peanut butter cookies = 1
Probability of drawing chocolate chip cookies from plate 1 :
Probability =( number of required outcome / Total possible outcomes)
P(chocolate chip) = 4 / 5
Plate 2:
Number of chocolate chip = 2
Number of peanut butter cookies = p
P(chocolate chip) = 2 / (2 + p)
Probability of drawing chocolate chip from plate 1 and then plate 2 = 8/ 25
(4/5) * 2/(2+p) = 8/ 25
8 / (10 + 5p) = 8/ 25
8(10 + 5p) = 8 * 25
80 + 40p = 200
40p = 200 - 80
40p = 120
p = 3
