What is the sum of 8+(–4/1/6)

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

3 0

Use the Order of operations but don't use pemdas...use GEMS :)

Paul [167]3 years ago

3 0

A1 =a1+(n-1)*d =8+(1-1)*6 =8
a2 =a1+(n-1)*d =8+(2-1)*6 =14
a3 =a1+(n-1)*d =8+(3-1)*6 =20
a4 =a1+(n-1)*d =8+(4-1)*6 =26
a5 =a1+(n-1)*d =8+(5-1)*6 =32
a6 =a1+(n-1)*d =8+(6-1)*6 =38
a7 =a1+(n-1)*d =8+(7-1)*6 =44
a8 =a1+(n-1)*d =8+(8-1)*6 =50
a9 =a1+(n-1)*d =8+(9-1)*6 =56
a10 =a1+(n-1)*d =8+(10-1)*6 =62
a11 =a1+(n-1)*d =8+(11-1)*6 =68
a12 =a1+(n-1)*d =8+(12-1)*6 =74
a13 =a1+(n-1)*d =8+(13-1)*6 =80
a14 =a1+(n-1)*d =8+(14-1)*6 =86
a15 =a1+(n-1)*d =8+(15-1)*6 =92
a16 =a1+(n-1)*d =8+(16-1)*6 =98
a17 =a1+(n-1)*d =8+(17-1)*6 =104
a18 =a1+(n-1)*d =8+(18-1)*6 =110
a19 =a1+(n-1)*d =8+(19-1)*6 =116
a20 =a1+(n-1)*d =8+(20-1)*6 =122
a21 =a1+(n-1)*d =8+(21-1)*6 =128
a22 =a1+(n-1)*d =8+(22-1)*6 =134
a23 =a1+(n-1)*d =8+(23-1)*6 =140
a24 =a1+(n-1)*d =8+(24-1)*6 =146
a25 =a1+(n-1)*d =8+(25-1)*6 =152
a26 =a1+(n-1)*d =8+(26-1)*6 =158
a27 =a1+(n-1)*d =8+(27-1)*6 =164
a28 =a1+(n-1)*d =8+(28-1)*6 =170
a29 =a1+(n-1)*d =8+(29-1)*6 =176
a30 =a1+(n-1)*d =8+(30-1)*6 =182
a31 =a1+(n-1)*d =8+(31-1)*6 =188
a32 =a1+(n-1)*d =8+(32-1)*6 =194

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