Question 1
Volume of cone = [tex] \frac{1}{3} \pi r^{2} h= \frac{1}{3} \pi (1.5)^{2} ( 5)=11.775[/tex]
Question 2
Scale factor of length = 0.75
Scale factor of area = 0.75² =0.5625
Area of original rectangle = 384
Area of the resized rectangle = 216
Question 3
Radius = 16
Angle = 45°
Area = [tex] \frac{45}{360}[/tex]×[tex] \pi (16)^{2} [/tex]=[tex]32 \pi [/tex]
Question 4
Using the trigonometry ratio
[tex]tan(x)= \frac{opposite}{adjacent} [/tex]
[tex]tan (56)= \frac{opposite}{26} [/tex]
opposite side = [tex]26tan(56)=39[/tex] ft to the nearest ten
Question 5
Circle equation is given by [tex](x-a)^{2}+ (y-b)^{2}= r^{2} [/tex]
[tex] x^{2} +10x+ y^{2}+12y+50=0 [/tex], using the completing the square
[tex] (x+5)^{2}-25+ (y+6)^{2}-36+52=0 [/tex]
[tex] (x+5)^{2}+ (y+6)^{2}=-52+36+25 [/tex]
[tex] (x+5)^{2}+ (y+6)^{2}=9 [/tex]
Question 6
First, we need to find the distance between [tex]( x_{0}, y_{0} ) [/tex] and the focus. [tex]( x_{0}, y_{0} ) [/tex] is any point on the parabola. We also need to find the distance between [tex]( x_{0}, y_{0} ) [/tex] and the directrix
The distance between [tex]( x_{0}, y_{0} ) [/tex] and (-2,4) is given
[tex] \sqrt{( x_{0}+2) ^{2}+ ( y_{0}-4)^{2} } [/tex]
The distance between [tex]( x_{0}, y_{0} ) [/tex] and y=6 is given [tex]| y_{0}-6| [/tex]
Equating the two expressions gives
[tex] \sqrt{( x_{0}+2) ^{2}+ ( y_{0}-4)^{2} } [/tex] = [tex]| y_{0}-6| [/tex]
Squaring both sides give
[tex](x_{0}+2) ^{2}+ ( y_{0}-4)^{2}= (y_{0}-6)^{2} [/tex]
Expand and simplify
[tex] x_{0} ^{2}+4 x_{0} +4 y_{0}-16=0 [/tex]
[tex] y_{0}= \frac{x_{0} }{4}- x_{0}+16 [/tex]
Question 7
Total BTU of 15 people on the bus = 15×750 = 11250 BTU
The volume of the bus = 38×8.5×6.25=2018.75 cubic foot
BTU per cubic foot = 11250 ÷ 2018.75 = 5.57
Question 8
Density = Mass/Volume = 120÷(7³) = 0.35 (two decimal places)
Diagram 1
HIJF is a quadrilateral. Angle IHF and IJF makes right-angle to the radius. It leaves us with angle HJF remain unknown.
Angle HJF = 180°-101° = 79°
Length of arc HJ = [tex] \frac{79}{360}*281=61.7 [/tex] (round 1 dp)
Diagram 2
By using the trigonometry ratio
[tex]sin(x)= \frac{opposite}{hypotenuse} [/tex]
[tex]sin(x)= \frac{15}{115} [/tex]
[tex]sin^{1}( \frac{15}{115})=x [/tex]
[tex]x=7.5[/tex]
Diagram 3
Area of triangle = [tex] \frac{1}{2}(8)(13)sin(33)=28.32 [/tex]
Diagram 4
Let O be the centre of the circle
Angle AOC = 25×2=50°
