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Circles that share a common center but no other common points are called:
1 answer:
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Answer:
71 remainder 80
Step-by-step explanation:
See the picture attached with letters to better understand the problem
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units
in the right triangle ABD
applying the Pythagoras theorem
8²=AD²+(x+4)²-----> AD²=64-(x+4)²------> equation 1
in the right triangle ADC
applying the Pythagoras theorem
12²=AD²+(2x+1)²-----> AD²=144-(2x+1)²------> equation 2
equate equation 1 and equation 2
64-(x+4)²=144-(2x+1)²-------> 64-[x²+8x+16]=144-[4x²+4x+1]
64-x²-8x-16=144-4x²-4x-1
4x²-x²-8x+4x=144-1-64+16
3x²-4x=95
3x²-4x-95=0
using a graph tool----> to resolve the second order equation
see the attached figure N 2
the solution is
x=6.33
the answer is
x=6.33 units