solve each polynomial equation
1.x3 -3x2 + 3x -1
2. x3- 3x2 +3x -1
brainliest ko tama ang sagot
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solve each polynomial equation
1.x3 -3x2 + 3x -1
2. x3- 3x2 +3x -1
brainliest ko tama ang sagot
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Answer:
✏️POLYNOMIAL EQUATION
==============================
Directions: Find S, given the following conditions:
\begin{gathered} \begin{aligned} & \bold{Formula:} \\ & \boxed{S_n = \frac{n}{2}\big[ 2a_1 + d(n-1) \big]} \end{aligned} \end{gathered}
Formula:
S
n
=
2
n
[2a
1
+d(n−1)]
#1: \sf a_1 = 2 \:;\:\: d = 6 \:;\:\: n = 18a
1
=2;d=6;n=18
\begin{gathered} S_{18} = \frac{18}{2} \big[ 2(2) + 6(18-1) \big] \\ \end{gathered}
S
18
=
2
18
[2(2)+6(18−1)]
\begin{gathered} S_{18} = \frac{18}{2} \big[ 2(2) + 6(17) \big] \\ \end{gathered}
S
18
=
2
18
[2(2)+6(17)]
S_{18} = 9 \big[4 + 102\big]S
18
=9[4+102]
S_{18} = 9 \big[106\big]S
18
=9[106]
S_{18} = 954S
18
=954
\therefore∴ The sum of the first 18 terms of the given arithmetic series is...
\Large \underline{\boxed{\tt \purple{\,954\,}}}
954
\:
#2: \sf a_1 = 5 \:;\:\: d = 9 \:;\:\: n = 20a
1
=5;d=9;n=20
\begin{gathered} S_{20} = \frac{20}{2} \big[ 2(5) + 9(20-1) \big] \\ \end{gathered}
S
20
=
2
20
[2(5)+9(20−1)]
\begin{gathered} S_{20} = \frac{20}{2} \big[ 2(5) + 9(19) \big] \\ \end{gathered}
S
20
=
2
20
[2(5)+9(19)]
S_{20} = 10 \big[10 + 171\big]S
20
=10[10+171]
S_{20} = 10 \big[181\big]S
20
=10[181]
S_{20} = 1,\!810S
20
=1,810
\therefore∴ The sum of the first 20 terms of the given arithmetic series is...
\Large \underline{\boxed{\tt \purple{\,1,\!810\,}}}
1,810
\:
#3: \sf a_1 = 2 \:;\:\: d = 6 \:;\:\: n = 10a
1
=2;d=6;n=10
\begin{gathered} S_{10} = \frac{10}{2} \big[ 2(2) + 6(10-1) \big] \\ \end{gathered}
S
10
=
2
10
[2(2)+6(10−1)]
\begin{gathered} S_{10} = \frac{10}{2} \big[ 2(2) + 6(9) \big] \\ \end{gathered}
S
10
=
2
10
[2(2)+6(9)]
S_{10} = 5 \big[4 + 54\big]S
10
=5[4+54]
S_{10} = 5 \big[58\big]S
10
=5[58]
S_{10} = 290S
10
=290
\therefore∴ The sum of the first 10 terms of the given arithmetic series is...
\Large \underline{\boxed{\tt \purple{\,290\,}}}
290
\:
#4: \sf a_1 = 4 \:;\:\: d = 3 \:;\:\: n = 12a
1
=4;d=3;n=12
\begin{gathered} S_{12} = \frac{12}{2} \big[ 2(4) + 3(12-1) \big] \\ \end{gathered}
S
12
=
2
12
[2(4)+3(12−1)]
\begin{gathered} S_{12} = \frac{12}{2} \big[ 2(4) + 3(11) \big] \\ \end{gathered}
S
12
=
2
12
[2(4)+3(11)]
S_{12} = 6 \big[8 + 33\big]S
12
=6[8+33]
S_{12} = 6 \big[41\big]S
12
=6[41]
S_{12} = 246S
12
=246
\therefore∴ The sum of the first 12 terms of the given arithmetic series is...
\Large \underline{\boxed{\tt \purple{\,246\,}}}
246
\:
#5: \sf a_1 = \text-3 \:;\:\: d = 2 \:;\:\: n = 10a
1
=-3;d=2;n=10
\begin{gathered} S_{10} = \frac{10}{2} \big[ 2(\text-3) + 2(10-1) \big] \\ \end{gathered}
S
10
=
2
10
[2(-3)+2(10−1)]
\begin{gathered} S_{10} = \frac{10}{2} \big[ 2(\text-3) + 2(9) \big] \\ \end{gathered}
S
10
=
2
10
[2(-3)+2(9)]
S_{10} = 5 \big[\text-6 + 18\big]S
10
=5[-6+18]
S_{10} = 5 \big[12\big]S
10
=5[12]
S_{10} = 60S
10
=60
\therefore∴ The sum of the first 1p terms of the given arithmetic series is...
\Large \underline{\boxed{\tt \purple{\,60\,}}}
60
==============================
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