9514 1404 393
Answer:
945x^3·y^16
Step-by-step explanation:
The 5th term will be ...
C(7,4)(3x)^3(y^4)^4 = 35·27·x^3·y^(4·4)
= 945x^3·y^16
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The coefficients of the expansion are C(n, k) where n is the exponent of the binomial, and k+1 is the term number. This is n!/(k!(n-k)!). The general (k-th) term of the expansion (a+b)^n is C(n, k)a^(n-k)b^(k) where k ranges from 0 to n.
Answer:
B. $2862
Step-by-step explanation:
Using n=5 in the given equation, we get ...
A(5) = 2700 + (5-1)(.015·2700) = 2700 +4(40.50)
A(5) = 2862.00
In year 5, you will have $2862 in the account.
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<em>Comment on the given equation</em>
The given equation tells you the amount in the account at the <em>beginning</em> of the year, before it earns any interest. Since that is the equation given, we presume that is the answer desired. In most "account balance" problems, you are interested in the amount at the <em>end</em> of the interest-earning period.
The answer is left two places.
For example 5.% would be 0.05 as a decimal.
Answer:
221 remainder 1
Step-by-step explanation:
885/4 is equal 221.25
This means that the quotient is 221, and when multiplying 221 by four gets us 884. 885 - 884 has a remainder of 1
Answer:
YXV- WZV- Congruent-Vertical- AA
Step-by-step explanation:
