Answer to Question #160227 in Algebra for Smayya Ruganyira

Question

Using a ruler, pencil and pair of compasses only, construct a triangle ABC such that length AB=8.7 cm, AC=10.6 cm and angle BAC=600. Find the length BC. (You must show all construction lines).

Given the function f(x)=x+4 : -x+4 ;

a. Find f(-¼)

b. Calculate the value of x for which x is not defined.

Solve for x, over set of real numbers: x²-x-90=0

Line A is parallel to line B. line b passes through points (4,5) and (1, -4).Find the equation of line A if it passes through (0, -1).

Given that f(x)= 2x+1 and g(x)= x2-9, find the value of x if gf(x)=0.

The shortest side of a triangle is twice the length of the shortest side. The third side is 7 cm less than the longest side and the perimeter of the triangle is 78 cm.

If 5/√5 + √20=a √5 ,find the value of a.

A factory, a worker’s wages for a 40 hours week is 120,000Frw. She is paid 10% of her weekly wages for every 2 hours that she works overtime. At the end of certain week she received 216,000Frw. Calculate the number of hours overtime that she worked.

A car was bought for 33,000,000Frw. The value of a car depreciated each year by 15% of its value at the beginning of the year. Calculate the value of the car after 3 years.

a) given the function p(x)= 6x3+35x2+19x-30.

b) prove that -5 is zero of p(x) and hence factorize it completely.

c. Find the values of x for which p(x)=0.

Given that a=3, b=-2 and c=4, find the value of : ab²-bc+ac.

Find the value of n : 41ten=45n

The interest on a loan is 24% per annun. How much is loan that bears interest of 6,000Frw after 1 year?

If Y inversely proportional to X and Y=40 when X=3, find Y when X=2.5

Solve the following simultaneous equations:

8x+y=21

5x-4y=10

Find the equation of line with gradient 5 and passing through the point (1,9).

Solve the following inequality:

1/3x-(x+2)>2

In a class of 50 students,40 likes mathematics and 25 likes science. Some students like both subjects and 2 do not like any of the subjects. How many students like both mathematics and science?

Answer 1

1.)

Step 1 : Draw a line segment AB of length 8.7cm.With the compass on point A, draw an arc across AB and up over above point A.

Step 2 : Without changing the compass width, move the compass to the point where the arc crosses AB, and draw an arc that crosses the first one.

Step 3 : Join point A to the point where the two arcs meet (point P) and extend the line to a point X .Now we get the angle BAX = 60°.

Step 4 : Now measure the compass width of length 10.6 cm using the scale and put the needle on point A and cut an arc on line AX and that intersecting point is C with the length AC of 10.6 cm.

Step 5 : Now join the point C to the point B. This gives us a triangle ABC in which angle BAC = $60\degree$ and AB = 8.7 cm and AC = 10.6 cm.

For length BC we can easily measure with the Scale or can be find mathematically by cosine rule:

$cosA=\frac{b^2+c^2-a^2}{2bc}$

where A,B,C are angles of triangle whose opposite side lengths are as a,b,c respectively. Thus

$cos 60\degree=\frac{(10.6)^2+(8.7)^2-a^2}{2*(10.6)*(8.7)}=\frac{1}{2}$

112.36+75.69$-a^2$ = 92.22

a = $\sqrt{95.83}$ = 9.8 cm

thus side BC = 9.8 cm

2.)

a)

$f(\frac{-1}{4})=\frac{\frac{-1}{4}+4}{\frac{-(-1)}{4}+4}=\frac{15}{17}=0.882$

b)

$f(x) = \frac{x+4}{-x+4}$ , this function can be define for any value of x except x= 4, as in denominator zero is created due to which function get undefined.

f(4) $=\frac{8}{0}=$ infinity value (undefined)

3.)

x²-x-90 = 0

$x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a}$

$x= \frac{-(-1)\pm\sqrt{(-1)^2-4*1*(-90)}}{2*1}=\frac{1\pm\sqrt{361}}{2}$

$x=10$ or $-9$

4.)

Equation of line B pass through points (4,5) and (1,-4) is

$(y-y_1)=\frac{y_2-y_1}{x_2-x_1}*(x-x_1)$

$(y-5)=\frac{-4-5}{1-4}*(x-4)$

$y=3x-7$

now let the equation of line A is $y=mx+c$

given that line A is parallel to line B so slope of both lines are equal and slope is m = 3.

Also the line pass through the point (0,-1), thus these point also satisfy the equation of line A.

-1 = 3*(0)+c

c = -1

hence, the equation of line A is y = 3x-1 .

5.)

given that: f(x)= 2x+1 and g(x) = x²-9 and g(f(x)) = 0

g(f(x)) = (2x+1)² - 9 = 0

4x² + 4x - 8 = 0

$x=\frac{-4\pm\sqrt{4^2-4*4*(-8)}}{2*4}=\frac{-4\pm\sqrt{144}}{8}$

x = -2,1

6.)

Let the length of shortest side be x cm.

According to the given information,

Longest side = 2 * Shortest side = 2x cm

And third side = longest side - 7 = (2x-7) cm

Perimeter of triangle = x + 2x + (2x-7) = (5x - 7) cm

But it is given that,

perimeter = 78cm = 5x - 7

78+7 = 5x

x = 17 cm

thus, the sides of the triangle are : 17 cm, 27 cm, 34 cm.

7.)

$\frac{5}{\sqrt{5}+\sqrt{20}}=a*\sqrt{5}$

$\frac{\sqrt{5}*{\sqrt{5}}}{\sqrt{5}*(1+{\sqrt{}4})}=a*{\sqrt{5}}$

$\frac{\sqrt{5}}{1+2}=a*{\sqrt{5}}$

$a= \frac{1}{3}$

8.)

A worker’s wages for a 40 hours week is = 120,000Frw

For every 2 hours overtime she paid = 10%(weekly wages) =$\frac{10}{100}*(120000)=12000Frw$

so for one hour overtime she paid =$\frac{12000}{2}=6000Frw$

but she get weekly wages = 216000Frw

The extra wage come from the overtime hours = 216000-120000 = 96000Frw

so the income from the overtime in a week =$\frac{96000}{6000}=16 hours$

9.)

Price of car = 33,000,000Frw

value of a car depreciated each year by 15% of its value at the beginning of the year$= \frac{15}{100}*(33,000,000)=4,950,000Frw$

value of car after first year completion = 33,000,000 - 4,950,000 = 28,050,000Frw

after 2nd year car value depreciated is $= \frac{15}{100}*(28,050,000)=4,207,500 Frw$

value of car after second year completion = 28,050,000 - 4,207,500 = 23,842,500Frw

after 3rd year car value depreciated is $= \frac{15}{100}*(23,842,500)=3,576,375 Frw$

value of car after third year completion = 23,842,500 - 3,576,375 = 20,266,125Frw

thus the value of car after 3 years = 20,266,125Frw

10.)

a)

given equation is p(x)$= 6x^3+35x^2+19x-30$

b)

for x = -5,

$p(-5)=6(-5)^3+35(-5)^2+19(-5)-30$

$p(-5)=-750+875-95-30=0$

hence, p(-5) = 0 proved.

c)

to factories the above equation, firstly find one solution of this equation and that is x = -5

so x+5 =0 is a factor of the given equation. On dividing the equation by (x+5), we get$=\frac{6x^3+35x^2+19x-30}{x+5}= 6x^2+5x-6$

so the above given equation can be written as $(x+5)(6x^2+5x-6)$

further the equation $6x^2+5x-6$ can be factories as

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-5\pm\sqrt{5^2-4*6*(-6)}}{2*6}=\frac{-5\pm\sqrt{169}}{12}=\frac{-5\pm13}{12}$

$x=\frac{-3}{2}or\frac{2}{3}$

so the solutions of equation (6x²+5x-6) are: x= -3/2 i.e. (2x+3) and x = 2/3 i.e. (3x-2)

therefore equation 6x²+5x-6 can be written as (2x+3)(3x-2).

so factorization of $6x^3+35x^2+19x-30$ is (x+5)(2x+3)(3x-2) .

the values of x are :

x+5 = 0, x=-5,

2x+3=0, x = -3/2

3x-2 = 0, x= 2/3

11.)

a = 3, b = -2, c = 4

ab²-bc+ac = 3*(-2)² - (-2)*4 + 3*4 = 12+8+12 = 32

12.)

$45_n = 41_{ten}$

45 with base n = 41 with base 10

$4*(n)^1 + 5*(n)^0 = 4*(10)^1 + 1*(10)^0$

4n + 5*1 = 40 + 1

4n + 5 = 41

n = 9

13.)

given that :

rate of interest = 24% per annum

time = 1 year

interest = 6000Frw

interest = (principle amount * rate * time)/100

6000 = (p * 24 * 1)/100

p = 600000/24

p(loan amount) = 25000Frw

14.)

$y\ltimes\frac{1}{x}$

$y=\frac{k}{x}$ , where k = constant value

y = 40, when x = 3

$40=\frac{k}{3}$

k = 120

for x = 2.5, $y=\frac{120}{2.5}$

y = 48

15.)

8x+y=21 (equation 1)

5x-4y=10 (equation 2)

multiply the equation 1 by 4 and add it to the equation 2.

4*(8x+y) + (5x-4y) = 84+10

37x = 94

x = 2.54

16.)

equation of line having , gradient (slope 'm') of line = 5 and pass through point (1,9) is

y = mx +c

9 = 5*1+c or c=4

so, y = 5x+4

17.)

(1/3x)-(x+2)>2

$\frac{1}{3x}-x-2\gt2$

$\frac{1}{3x}\gt x+4$

$3x^2+12x-1\lt 0$

$x=\frac{-12\pm\sqrt{12^2-4*3*(-1)}}{2*3}=\frac{-12\pm\sqrt{156}}{6}$

x = 0.082

$x-0.082\lt0$

$x\lt0.082$

or

x = -4.082

$x+4.082\lt0$

$x\lt-4.082$

18.)

total students = 50

students like math = 40

students like science = 25

let students like both math and science = x

students don't like any of the subjects = 2

total students = like math + like science - (like both subject) + don't like any subjects

50 = 40+25-x+2

x = 17

so, the total number of students who likes the both subjects are 17.