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Sagot :
[tex]Equation; \\ \\ x+1+x+3+x+5+x+7=152 \\ \\ 4x+16=152 \\ \\ 4x=152-16 \\ \\ 4x=136 \\ \\ \frac{4x}{4}= \frac{136}{4}\ \ \ \ \ \ |\ Divide\ both\ terms\ by\ 4\ so\ we\ could\ get\ the\ value\ of\ x \\ \\ \boxed{x=34} [/tex]
[tex]Substitute\ the\ given\ equation\ where\ x\to\ 34 \\ \\ \\ x+1\to34+1=\boxed{35} \\ \\ x+3\to34+3=\boxed{37} \\ \\ x+5\to34+5=\boxed{39} \\ \\ x+7\to34+7=\boxed{41} \\ \\ \\ Therefore,\ the\ four\ consecutive\ odd\ numbers\ that\ is\ equal\ to\ 152 \\ is\ \underline{35},\ \underline{37},\ \underline{39}\ and\ \underline{41}... \\ \\ Hope it Helps:) \\ Domini[/tex]
[tex]Substitute\ the\ given\ equation\ where\ x\to\ 34 \\ \\ \\ x+1\to34+1=\boxed{35} \\ \\ x+3\to34+3=\boxed{37} \\ \\ x+5\to34+5=\boxed{39} \\ \\ x+7\to34+7=\boxed{41} \\ \\ \\ Therefore,\ the\ four\ consecutive\ odd\ numbers\ that\ is\ equal\ to\ 152 \\ is\ \underline{35},\ \underline{37},\ \underline{39}\ and\ \underline{41}... \\ \\ Hope it Helps:) \\ Domini[/tex]
I'll solve it using the RESAC Form.
R: Let x = first number
x + 2 = second number
x + 4 = third number
x + 6 = fourth number
E : x + x + 2 + x + 4 + x + 6 =152
S : x + x + 2 + x + 4 + x + 6 = 152
4x + 12 = 152
[tex] \frac{4x}{4} = \frac{140}{4} [/tex]
x = 35 ; x + 2 = 37 ; x + 4 = 39; x + 6 = 41
A : The four consecutive odd numbers that make up the sum of 152 are 35, 37, 39 and 41.
R: Let x = first number
x + 2 = second number
x + 4 = third number
x + 6 = fourth number
E : x + x + 2 + x + 4 + x + 6 =152
S : x + x + 2 + x + 4 + x + 6 = 152
4x + 12 = 152
[tex] \frac{4x}{4} = \frac{140}{4} [/tex]
x = 35 ; x + 2 = 37 ; x + 4 = 39; x + 6 = 41
A : The four consecutive odd numbers that make up the sum of 152 are 35, 37, 39 and 41.
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