Answer:
i think the answer is B but im not sure sorry if it is not
Step-by-step explanation:
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
a. 4 – commutative property
b. 5 – commutative property
c. 0 – identity property
d. 4 – associative property
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The commutative property lets you swap the order: (a) + (b) = (b) + (a).
The associative property lets you change the grouping: (a+b)+c = a+(b+c).
The identity property lets you add 0 without changing anything: (a) +0 = (a).
Here let us first write the value of each number:
6 thousands means 6*1000=6000.
9 hundreds means 9*100=900.
12 tens means 12*10 =120
10 ones means 10*1 =10
3 tenths means 3/10 =0.3
Now we just have to add all these values to get the final number.
adding 6000 + 900 +120 +10 +0.3 = 7030.3
So the final number is 7030.3
Answer: C
Step-by-step explanation: This is actually very simple, all you have to do is type in a calculator 25 divided by 6. After that you will be given 4.1666666 repeating. So your answer is 4.16 with a repeating 6. Hope this helps!
