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son4ous [18]
4 years ago
5

Solve 1)3x-7=35? 2)22=7-3a 3)-d+7=3 4)2x+23=49 5)-c+2=5

Mathematics
2 answers:
kondaur [170]4 years ago
8 0
1)\\3x-7=35\\3x-7+7=35+7\\3x=42\\3x=3\cdot14\ /:3\\x=14\ \ \ \leftarrow\ \ \ Answer\\\\
2)\\22=7-3a\\22-22+3a=7-3a-22+3a\\3a=-15\\3a=3\cdot(-5)\ /:3\\a=-5\ \ \ \leftarrow\ \ \ Answer\\\\3)\\-d+7=3\\-d+7-7=3-7\\-d=-4\\d\cdot(-1)=4\cdot(-1)\ /:(-1)\\d=4\ \ \ \leftarrow\ \ \ Answer\\\\

4)\\2x+23=49\\  2x+23-23=49-23\\2x=26\\2\cdot x=2\cdot 13\ /:2\\x=13  \ \ \ \leftarrow\ \ \ Answer\\\\ 
5)-c+2=5\\  -c+2-2=5-2\\ -c=3\\ c\cdot(-1)=-3\cdot(-1)\ /:(-1)\\c=-3 \ \ \ \leftarrow\ \ \ Answer
astra-53 [7]4 years ago
7 0
1)\ 3x-7=35\ \ \ |add\ 7\ to\ both\ sides\\\\3x=42\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=14}\\-------------------------\\2)\ 22=7-3a\\\\7-3a=22\ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-3a=15\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\\\\\boxed{a=-5}\\--------------------------

3)\ -d+7=3\ \ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-d=-4\\\\\boxed{d=4}\\---------------------------\\\\4)\ 2x+23=49\ \ \ \ \ \ |subtract\ 23\ from\ both\ sides\\\\2x=26\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=13}\\---------------------------\\5)\ -c+2=5\ \ \ \ \ \ |subtract\ 2\ from\ both\ sides\\\\-c=3\\\\\boxed{c=-3}
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First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

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