Rent paid during 1995 was N2,000 while rent paid at 31st December, 1995 was
N2,400
N2,200
N2,000
N1,800
N200
Correct answer is D
No explanation has been provided for this answer.
The double entries for refund of unsuccessful application monies are, debit
Application for shares account, credit bank account
Bank account, credit application for shares account
Allotment account, credit application for shares account
Allotment account, credit bank account
Cash account, credit bank account
Correct answer is A
No explanation has been provided for this answer.
If \((x - 5)\) is a factor of \(x^3 - 4x^2 - 11x + 30\), find the remaining factors.
\((x + 3) and (x - 2)\)
\((x - 3) and (x + 2)\)
\((x - 3) and (x - 2)\)
\((x + 3) and (x + 2)\)
Correct answer is A
(x - 5) is a factor of \(x^3 - 4x^2 - 11x + 30\). To find the remaining factors, let's draw out \((x - 5)\) from the parent expression.
\(x^3 - 4x^2 - 11x + 30 = x^3 - 5x^2 + x^2 - 5x - 6x + 30\)
\(= x^2(x - 5) + x(x - 5) - 6(x - 5) = (x - 5)(x^2 + x - 6)\)
∴ To find the remaining factors, we factorize \((x2 + x - 6)\)
\(x^2 + x - 6 = x^2 + 3x - 2x - 6\)
\(= x(x + 3) - 2(x + 3) = (x + 3)(x - 2)\)
∴ The other two factors are \((x + 3) and (x - 2)\)
ALTERNATIVELY
\(∴ x^2 + x - 6 = (x + 3) and (x - 2)\)
In how many ways can four Mathematicians be selected from six ?
90
60
15
360
Correct answer is C
\(=^6C_4\)
\(=\frac{6!}{(4!\times2!)}\)
\(=\frac{6\times5}{2\times1}\)
= 15
\(-\frac{896x^6}{9}\)
\(-\frac{896x^5}{9}\)
\(-\frac{896x^5}{27}\)
\(-\frac{896x^6}{27}\)
Correct answer is C
rth term of a binomial expansion =\(^nC_r-1 a^{n-(r-1)}b^{r-1}\)
\(n = 10,r = 6 \therefore r-1=5\)
6th term =\(^{10}C_5 1^{10} - 5 (-\frac{2}{3}x)^5\)
\(=252*1*-\frac{32x^5}{243}=-\frac{896x^5}{27}\)