Answer to Question #160240 in Calculus for Smayya Ruganyira

Question

The vertices of a triangle “STV” are S(0, 2), T(0,5) AND V(0, 3). In the same Cartesian plane, draw: 

a) The triangle STV.              

b) (i) the triangle S’T’V’, image of triangle STV under reflection in y-axis.  

(ii) the triangle S”T”V”, image of triangle STV under a rotation about the origin through 90° 

(iii) the triangle S’”T’”V”, image of triangle STV under translation , T=(1) .

3

A school has a teaching staff of 22 teachers. 8 of them teach Mathematics, 7 teach Physics and 4 teach Chemistry. 3 teach both Mathematics and Physics, none teaches Mathematics and Chemistry. No teacher teaches all the three subjects. The number of teachers who teach Physics and Chemistry is equal to the number of teachers who teach Chemistry but not Physics.

(a) Represent the data on a Venn diagram. (b) Find the number of teachers who teach; (i) Mathematics only.

(ii) Physics only.     

(iii) None of the three subjects.

Given that, n(A∪B) = 29, n(A) = 21, n(B) = 17, n(A∩B) = x.

a. Write down in terms of the elements of each part.

b. Form an equation and hence find the value of x.

 

Carry out the following in base six        

(a) 115 + 251 + 251       

(b) 53412 - 34125           

(c) 123 × 54

If A stands for 10 and B stands for eleven, perform the following duodecimal calculations:      

(a) 59A + AB

(b) 4A + AB + 9      

(c) 10 × 54

(d) 45B – A1      

(e) 11 × 7 – 8

(f) 32 + 6B         

(g) 159A – 6BA

(a) Draw the region given by the inequalities x > 0, 5x + 4y < 32, x +2y>10.

(b) State the coordinates of the vertices of the region in (a).

A farmer makes a profit of x FRW on each of the (x + 5) eggs her hen lays. If her total profit was 84,000 FRW, find the number of eggs the hen lays.

Given that the line y = 3x + a passes through (1, 4), find the value of a.

Draw the graph of y = 2 + 2x – x2 for values of x from –2 to 4. From the graph, find:           

(a) the maximum value of 2 + 2x – x²         

(b) the value of x for which y is greatest     

(c) the range of values of x for which y is positive            

(d) the axis of symmetry

Two tall buildings A and B are 40 m apart. From foot A, the angle of elevation of the top of B is 60°. From the top of A, the angle of depression of the top of B is 30°. Find the heights of A and B, to the nearest 1 m.     

Find the value of the side marked.

Jacob and Bernard stand on one side of a tower and in a straight line with the tower. They each use a clinometer and determine the angle of elevation of the top the tower as 30° and 60° respectively. If their distance apart is 100 m, find the height of the tower.

Two girls one east and the other west of a tower,measure the angles of elevation of the top its spire as 28° and 37°. If the top of the spire is 120m high,how far apart are the girls?

Answer 1

a, b

(i) We need to get the new vertices of the triangle under reflection in y-axis. The new coordinate can be gotten by

$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$

So, let the vertices of the triangle STV be represented as a single matrix as $\begin{pmatrix} 0 & 0 & 0 \\ 2 & 5 & 3 \end{pmatrix}$

The coordinates of the new image formed will be

$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0\\ 2 & 5 & 3 \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0\\ 2 & 5 & 3 \end{pmatrix}$

Therefore, we will have $S'(0,2), T'(0,5) \text{ and } V'(0,3).$

(ii) The vertices of the new image form under a rotation is given by

$\begin{pmatrix} x'\\ y' \end{pmatrix}=\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$

Here, we will have $\theta=90^0$ . Therefore, the new vertices will be;

$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0 \\ 2 & 5 & 3 \end{pmatrix}=\begin{pmatrix} -2 & -5 & -3\\ 0& 0 & 0 \end{pmatrix}$

So, we will have $S''(-2,0), T''(-5,0) \text{ and } V''(-3,0).$

(iii) We need to get the new coordinate of the translated image by adding the translation matrix $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ to the old coordinate. So, we will have that

$\begin{pmatrix} 0 & 0 & 0 \\ 2 & 5 & 3 \end{pmatrix}+\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 1 & 1 \\ 3 & 6& 4 \end{pmatrix}$

Therefore, we will have $S'''(1,3), T'''(1,6) \text{ and } V'''(1,4).$

From the diagram above, we can see that the vertices given is not for a triangle. All the points are on a straight line.

3. a)

b) (i)The number of mathematics teacher only $=n(M)-n(M \bigcap P) =8-3=5 \text{ teachers.}$

(ii) Number of physics teacher only $n(P)-n(M \bigcap P) -n(P \bigcap C) =7-3-2=2 \text{ teachers. }$

(III) None of the three subjects $=22-(5+3+2+2+2)=22-14=8 \text{ teachers.}$

4. a) $n(A \bigcup B)$ is the number of elements in A or B, $n(A)$ is the number of elements in A, $n(B)$ is the number of elements in B, $n(A \bigcap B)$ is the number of elements in both A and B.

b)

$n(A \bigcup B) =n(A)+n(B)-n(A \bigcap B) \\ 29=21+17-x\\ 29=38-x\\ x=9.$

5. Note that all answers are in base 6.

a) 1101

b) 15243

c) 12010

6. Note that all answers are duodecimals

a) 689 b) 146 c) 540 d) 37A e) 6B

f) A1 g) AA0

7. a

b) The coordinates of the vertices in (a) are (0,5),(0,8) and (4,3)

7.

$x(x+5)=84\\ x^2+5x-84=0\\ (x-7) (x+12)=0\\ x=7,x=-12$

Taking the positive value. So the number of eggs sold= 7+5=12eggs

8. Since the point is on the line, substitute the value of x and y into the line equation

$4=3+a\\ a=1$

9.

From the graph above, we can see that

a) The maximum value of the function y is 3

b) The value if x when y is the greatest is 1

c) The range if x when you is positive is (-0.73,2.73)

d) The axis if symmetry is (1,3)

10.

Let the height of A and B be H and h respectively.

By trigonometric ratio, we have that

$\frac{h}{40}=\tan 60°,h=40\tan 60°=69.28m ≈69m\\ H=h+40\tan 30°=40 \tan 60° + 40 \tan30°= 92.37≈92m$

11.

Let the distance between the Tower and Bernard be x and the height of the tower be h.

By trigonometric ratio, we have that

$\frac{h}{x}=\tan 60°, h= x\tan 60°.............(1)\\ \frac{h}{100+x}=\tan 30°, h=(100+x)\tan 30......(2)\\ \text{Equate the two equations }\\ x\tan 60=(100+x)\tan 30\\ x=50m\\ \text{Substitute the value of x into (1)}\\ h=50 \tan 60=86.60m$

12.

Let the distance of A and B from the tower be x and y respectively.

By trigonometric ratio, we have that

$\frac{120}{x}=\tan 37°,x=\frac{120}{\tan 37°}=159.25m\\ \frac{120}{y}=\tan 28°,y=\frac{120}{\tan 28°}=225.69m\\ AB=x+y\\ AB=159.25+225.69=384.94m$