Answer:
1. [tex]2^{3}[/tex] x [tex]3^{6}[/tex]
ii. 18
b. -12
2. nth = a + (n - 1)d
ii. k = 103
3. [tex]x^{o}[/tex] = [tex]27^{o}[/tex]
4. Volume = 25.025 [tex]m^{3}[/tex]
Step-by-step explanation:
1. Prime factors of 5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
= [tex]2^{3}[/tex] x [tex]3^{6}[/tex]
ii. [tex]\sqrt[3]{5832}[/tex] = 18
b. 2[tex]a^{2}[/tex] - 3[tex]b^{2}[/tex] + 3abc
a = 3, b = 2, c = -1
Thus,
2[tex](3)^{2}[/tex] - 3[tex](2)^{2}[/tex] + 3(3 x 2 x -1)
= 18 - 12 - 18
= -12
2. The required formula is;
nth = a + (n - 1)d
where: a is the first term, n is the number of terms in the sequence and d is the common difference.
b. kth term = 304, then
304 = -2 + (k - 1) 3
= -2 + 3k -3
= 3k - 5
3k = 304 + 5
3k = 309
k = [tex]\frac{309}{3}[/tex]
= 103
k = 103
3. Comparing <DAB and < EBC;
<DAB ≅ <EBC = [tex]x^{o}[/tex]
<BEC = [tex]90^{o}[/tex] (right angle property)
<C = [tex]63^{o}[/tex]
Thus,
<B + <C + <E = [tex]180^{o}[/tex] (sum of angles in a triangle)
[tex]x^{o}[/tex] + [tex]63^{o}[/tex] + [tex]90^{o}[/tex] = [tex]180^{o}[/tex]
[tex]x^{o}[/tex] + 153 = [tex]180^{o}[/tex]
[tex]x^{o}[/tex] = 180 - 153
= 27
[tex]x^{o}[/tex] = [tex]27^{o}[/tex]
4. Volume of a cylinder = [tex]\pi[/tex][tex]r^{2}[/tex]h
where r is the radius and h is the height of the cylinder.
Diameter = 3.5 m, h = 2.6 m
r = [tex]\frac{3.5}{2}[/tex]
r = [tex]\frac{7}{4}[/tex] m
So that,
Volume = [tex]\frac{22}{7}[/tex] x [tex](\frac{7}{4}) ^{2}[/tex] x 2.6
= [tex]\frac{22}{7}[/tex] x [tex]\frac{49}{16}[/tex] x [tex]\frac{13}{5}[/tex]
= 25[tex]\frac{1}{40}[/tex]
Volume = 25.025 [tex]m^{3}[/tex]
