Assignment About Mathematics

Mathematics Assignment:
Student Name:
Instructor Name:
Course Number:
8th January 2020

Q1)
csc θ= −12
7
1
sin θ =−12
7
sin θ=−7
12
7 12
θ
√ 95
√ 122−72= √ 95
Since θ lies in the fourth quadrant, its cosine and tangent will be positive and
negative respectively.
cos θ= √95
12 = 9.747
12
tanθ= −7
√ 95 = −7
9.747
tanθ= −7
9.747
sec θ= 1
cosθ = 1
9.747
12
= 12
9.747
sec θ= 12
9.747

cot θ= 1
tanθ = 1
−7
9.747
=−9.747
7
cot θ=−9.747
7
Q2a)
C ( x )=0.3456 x
C ( 35 )=0.3456 ( 35 )=12.096
C ( 35 )=12.096
b)
Domain of C ( x ) is from x=0 to x <75
Q3a)
The function g touches the x axis twice at points (-1, 0) and (5, 0). We may
therefore say that it is a quadratic equation in x with roots x=−1∧x=5.
Therefore we have
( x +1 ) ( x−5 )=0
¿ x2−4 x−5=0
Assuming that g takes the form g( x )=a x2 +bx+ c we get
g ( x )=x2−4 x−5
The function f is a curve with stationary point at (3 , 0).
Assuming that f takes the formf ( x)=a x2 +bx +c.
At the stationary point f ' ( x ) =2 ax +b=0
f ' ( 3 ) =2 a(3)+b=0
6 a+b=0 ...............................................i
At (3, 0)
f ( 3 )=a (3)2 +3 b+ c
0=9 a+3 b+ c .................................ii

f Also passes through (4 ,−1)
f ( 4 ) =a(4)2 +4 b+c
−1=16 a+4 b+c .................................iii
iii minus ii gives 7 a+ b=−1
7 a+ b=−1 ... ... ... ... ... ... . iv
Solving i and iv simultaneously by taking iv minus i we have
a=−1
7 a+ b=−1
7(−1)+b=−1
b=−1+7=6
−1=16 (−1)+ 4 (6)+ c
−1+16−24=c
−9=c
f ( x ) =a x2 +bx +c=−x2 +6 x−9
f ( x )=−x2 +6 x−9
f ( x )=−x2 +6 x−9
g ( x )=x2−4 x−5
g ( x )=x2−4 x−5
g ( 1 )=12−4 ( 1 )−5=−8
g ( 1 )=−8
f ( x )=−x2 +6 x−9
f ( 1 ) =−12+ 6 ( 1 )−9=−4
f ( 1 ) =−4
( f ° g ) ( 1 )=−8 ×−4=32
b)
g ( x )=x2−4 x−5