Which postulate or theorem proves that △ABC and △EDC are congruent?
1 answer:
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Y=((square root of 2)+1)
1) (y-1)^2 =2
so we can concludes that y-1 = squareroot of 2
so ((squareeroot of 2+1) - 1)^2 = 2
2)First we can convert 21÷2 fraction to decmial form which is 10.5
so our equation would be (c+1)^2= 10.5
so c+1 is the sqaureroot root of 10.5
so c would be the (square root of 10.5)-1
3)We can again convert those fractions to decimals so 3/2=1.5 and 71/4 =17.75
so our equation would be (e-1.5)^2=17.75
so (e-1.5) is the square root of 17.75
so e=((square root of 17.75)+1)
1) (y-1)^2 =2
so we can concludes that y-1 = squareroot of 2
so ((squareeroot of 2+1) - 1)^2 = 2
2)First we can convert 21÷2 fraction to decmial form which is 10.5
so our equation would be (c+1)^2= 10.5
so c+1 is the sqaureroot root of 10.5
so c would be the (square root of 10.5)-1
3)We can again convert those fractions to decimals so 3/2=1.5 and 71/4 =17.75
so our equation would be (e-1.5)^2=17.75
so (e-1.5) is the square root of 17.75
so e=((square root of 17.75)+1)