Question

Help with this problem guys
→no trolling please i really need it←

Answer 1

We are given the following lengths:

• XY (median) = 4m+3
• CO (upper base) = 2m-10
• PY (lower base) = 3m+37

We will solve it by using this formula:

`\boxed{\mathfrak{\purple {Median = (1)/(2)(lower \: base+upper \: base)}}}`

`\boxed{\mathcal{\purple { XY = (1)/(2)( CO+PY)}}}`

Lets solve it down now,

#16

`\green{ \sf4m + 3 = (1)/(2) (2m - 10 + 3m + 37)}`

`\red \implies \green{ \sf4m + 3 = (1)/(2) (5m+ 27)}`

`\red \implies \green{ \sf4m + 3 = (5)/(2) m + (27)/(2) }`

`\red \implies \green{ \sf8m + 6 = 5m + 27 }`

`\red \implies \green{ \sf8m + 6 - 5m = 27 }`

`\red \implies \green{ \sf8m - 5m= 27 - 6 }`

`\red \implies \green{ \sf3m = 21 }`

`\boxed{ \mathfrak{ \red{m = 7}}}`

#17

`\blue \longmapsto \orange{ \tt \: XY =4m + 3}`

`\blue \longmapsto \orange{ \tt \: XY =4(7) + 3}`

`\blue \longmapsto \orange{ \tt \: XY =28 + 3}`

`\boxed{ \underline{ \underline{ \blue \longmapsto \orange{ \tt \: XY =31}}}}`

#18

`\purple \longmapsto \pink{ \tt \: CO = 2m - 10}`

`\purple \longmapsto \pink{ \tt \: CO = 2(7) - 10}`

`\purple \longmapsto \pink{ \tt \: CO = 14 - 10}`

`\boxed{ \underline{ \underline{ \purple \longmapsto \pink{ \tt \:CO=4}}}}`

#19

`\purple \longmapsto \pink{ \tt \: PY = 3m + 37}`

`\purple \longmapsto \pink{ \tt \: PY = 3(7) + 37}`

`\purple \longmapsto \pink{ \tt \: PY = 21 + 37}`

`\boxed{ \underline{ \underline{ \purple \longmapsto \pink{ \tt \:PY=58}}}}`