Solve the following system:{12 x = 54 - 6 y | (equation 1)-17 x = -6 y - 62 | (equation 2)
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.
The perimeter of the first figure is 34 cm and the area is 64 cm².
The perimeter of the second figure is 38 cm and the area is 60 cm².
The perimeter of the third figure is 30 cm and the area is 36 cm².
The perimeter of the fourth figure is 72 cm and the area is 200 cm².
The perimeter of the fifth figure is 30 cm and the area is 36 cm².
To find the perimeter of each, we add the area of all sides. For the first figure, the missing sides are 1 cm and 6 cm. To find the area, we have two rectangles whose dimensions are 6x10 and 1x4.
For the second figure, the missing sides are 4 cm and 3 cm. To find the area, we have two rectangles whose dimensions are 4x12 and 3x4.
For the third figure, the missing sides are 3 cm, 3 cm and 8 cm. To find the area, we have two rectangles whose dimensions are 4x3 and 3x8.
For the fourth figure, the missing sides are 10 cm, 10 cm, 6 cm and 6 cm. To find the area, we have two squares whose dimensions are 10x10.
For the fifth figure, the missing sides are 3 cm and 9 cm. To find the area, we have two rectangles whose dimensions are 3x6 and 6x3.
Answer:
3(10)-7
Step-by-step explanation:
3(10)-7
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