what is the value of x so that bc is parallel to de
Detailed solutions and explanations to the 31 Sabbatum Maths subject level ane questions respective to those in sample 1, are presented.
- In the effigy beneath, AB and GE are parallel. Triangle ACD is isosceles with the lengths of CA and CD equal. The measures of angles FDE and GDH are lx� and 65� respectively. What is the measure of angle CAB?
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Solution
Since AB and GE are parallel, angles BAD and EDF are corresponding angles and therefore equal in size. Hence the size of angle BAD is 60�
Angles CDE and GDH are vertical angles and therefore equal in size. Hence the size of angle CDE is 65�. The size of angle CDA is given past
180� - (sixty� + 65�) = 55�
The size of angle CAB is given past
size of bending BAD - size of angle CAD = lx� - 55� = 5�
- If a, b and y are positive real numbers such that none of them is equal to 1 and b 2a + 6 = y two, which of these must be truthful?
Solution
Since a and b are positive
√(b 2a + vi) = b (1/ii) (2a + 6) = b a + three
Since y is positive
√(y 2) = y
Since b 2a + 6 = y ii, taking the square root of both sides of the equation, we obtain
y = b a + 3
- What is true about the graph of role f defined by
f(ten) = 4 + | | x | - 3 |
(I) Symmetric with respect to y-axis
(Two) Has no x intercepts
(Three) Has a y intercept at (0 , 7)
Solution
(I): Determine f(-x) and simplify it
f(-ten) = iv + | | - x | - 3 | = 4 + | | ten | - iii | since |-x| = |x|
Hence f(x) is even and its graph is symmetric with to the y axis.
(Two): Find whatever x intercepts by solving f(x) = 0
4 + | | x | - three | = 0
Add together -four to both sides of the higher up equation to obtain
| | x | - 3 | = - 4
The absolute value of an expression is never negative, hence the to a higher place equation has no solutions and therefore the graph of the given function has no ten intercepts.
(III): Detect the y intercept by finding f(0)
f(0) = 4 + | | 0 | - three | = seven
The graph of f has a y intercept at (0 , vii).
All 3 statements in (I), (II) and (III) are true.
- A number x is divided past 0.71 and the upshot decreased by 0.2. If the fifth root of the terminal issue equals -0.fifteen, what is the value of x rounded to the nearest thousandth?
Solution
x divided by 0.71 and the result decreased by 0.ii
x / 0.71 - 0.2
The fifth root of the final (in a higher place) result equals -0.fifteen
(ten / 0.71 - 0.2) one/five = -0.15
We need to solve the to a higher place equation in society to find x. Enhance both sides of the equation to the ability 5.
(x / 0.71 - 0.2) = (-0.15) 5
Solve for x.
10 = 0.71( (-0.15) 5 + 0.2 ) (appro.)= 0.14194608
Round to the nearest thousanth
ten (appro.)= 0.142
- Detect the constant k and so that the perpendicular bisector of the line segment with end points (k , 0) and (4 , half-dozen) has a slope of -3.
Solution
We first the slope m1 of the line through the given points
m1 = (6 - 0) / (4 - k) = 6 / (four - k)
The slope of the perpendicular to the line segment is equal to -3. For two lines to be perpendicular, the product of their slopes must be equal to - one. Hence
( - 3 )*(6 / (4 - k) ) = - 1
Solve the above for k.
-18 / (4 - thou) = - one
-18 = - iv + k
k = - 14
- Twoscore students are each registered to written report one or more than of three courses. 23 students are registered in one grade only and 14 students are registered in two courses but. How many are registered in all 3 courses?
Solution
We first utilise Venn diagram to represent the students registered in one course just which is represented in yellowish and the students who study two courses but which is represented in blue. .
If the total number of students studying ane, two or three courses is twoscore, then the number of students studying three course is given by
40 - 23 - fourteen = 3
- Which of the statements below are true almost the graphs of the equations x 2 + y 2 = 9 and (x + 6) 2 + y 2 = 9?
(I) Both graphs are circles.
(Ii) The two graphs bear on at one point.
(I) The two graphs accept ii distinct points of intersection.
Solution
The two equations are of the form.
(x - h) 2 + (y - k) two = r 2
which are equations of circles.
Let u.s. now find the points of intersection of the two circles by solving the system of equations.
(x + 6) two + y two = 9
x 2 + y 2 = nine
Subtract the left hand sides and right hand sides of the 2 equations as follows
(x + 6) 2 + y two - [ ten 2 + y two ] = 9 - 9
Which gives an equation with i variable simply
12x + 36 = 0
ten = - 3
Use equation x 2 + y 2 = 9 to discover y
3 ii + y 2 = 9
y = 0
The two circles have i point in common and therefore touch on at one point.
Only statement (I) and (Ii) are truthful.
- If the perimeter of a regular hexagon is equal to 6 a, then the area of this hexagon if given by:
Solution
A regular hexagon has 6 sides of equal size. Hence if the perimeter is 6 a, each side has a length equal to a. Too this regular hexagon may exist considered as fabricated up of 6 equilateral triangles of side a. .
We first discover the surface area A of an equilateral triangle of side a. The height h of the triangle is found using Pythagora'south theorem.
h = √( aii - (a/2)2 ) = a √3 / 2
The expanse A of the equilateral triangle of side a is given by
A = (1/2)* a √three / two * a = a2 √3 / 4
The given hexagon is made up of half-dozen equilateral triangle each of side a, hence the are of the hexagon is given by
vi * ( a2 √iii / 4 ) = 1.5 √3 a2
- AB is a bore to the circumvolve in the figure below and point C is on the circle. The measure of the bore is ten units and side AC has a length of 5 units. Find the measure of angle CBA
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Solution
Since AB is the diamter of the circumvolve, and then triangle ABC is a right triangle and
sin(angle CBA) = 5 / ten = 1/ii
Hence bending CBA has a measure of arcsin(1/2) = 30 degrees
- For what value of positive k does the equation have one solution only?
(10 + k) x = - iv
Solution
We first expand the left side and write the equation in standard form
x 2 + thou x + 4 = 0
The equation is quadratic and in guild to have one solution only, its discriminant D must be equal to 0. Permit us offset observe the discriminant D.
D = k 2 - 4 (1)(four) = m ii - 16
Solve D = yard 2 - 16 = 0 and select the positive solution
m = 4
- The average of the roots of a quadratic equation is equal to 3 and the difference of the roots is equal to 2. Which of these could be the equation?
Solution
Allow A and B be the roots of the equation. First their average is 3. Hence
(A + B) / 2 = three or A + B = 6
The deviation of the 2 roots is 2. Hence
A - B = 2
Solve the system of equations A + B = 6 and A - B = two to notice
A = 4 and B = two
A quadratic equation with roots 4 and two and leading coefficient one has the class
(ten - four)(ten - 2) = 0
Expand the left side of the equation.
x 2 - 6 x + 8 = 0
- If points One thousand, B(2 , 6) and C(4 , eight) are such that MBC is a correct triangle with hypotenuse BC, then the 3 points Thou, B and C are on a circle of radius Solution
Utilise the converse of Thales' theorem that states that if a right triangle is inscribed in a circle and so the hypotenuse will be a bore of the circle to describe triangle MBC as shown beneath..
If the bore is BC then the radius R is half the distnace BC. Hence
R = (1/2) √( (four - 2) 2 + (8 - 6) 2 ) = √2
- The volume of a rectangular solid is equal to grand m iii. If the length, width and summit are increased by fifty%, the book of the new rectangular solid is equal to
Solution
If L, Due west and H are the three dimensions of a rectangular solid so its volume V is given by
V = L * W * H
If L is increased by 50%, the new value of L will be
L + 50%L = 1.5 Fifty
Therefore increasing each dimension of the rectangle by 50% is the same every bit multiplying each of the iii dimensions by ane.v. Hence the volume of the new rectangular solid is
1.5L * 1.5W * 1.5H = (1.5) iii Fifty * W * H
= 3.375 * g = 3,375 m 3
- In the figure beneath angle ADE is right and BC and DE are parallel. The length of Air-conditioning is 5 and the length of BC is 4. Find tan(angle CED). .
Solution
Since BC is parallel to DE, triangles ABC and ADE are like and therefore triangle ABC is as well a right triangle. The corresponding angles of the two triangle are congruent and therefore equal in size. Hence
tan(CED) = tan(ACB)
Employ Pythagora's theorem to notice the length of side AB and discover tan(ACB)
AB = sqrt(52 - 452) = 3
tan(ACB) = 3/iv = tan(CED)
- What are the coordinates of the center of the circle that is entirely in quadrant Two and is tangent to the lines y = eight, y = 2 and x = -2?
Solution
Since the circumvolve is tangent to y = viii and y = 2, the y coordinate of the center is equal to
(viii + 2) / 2 = 5
and its radius is equal to
(viii - 2) / ii = 3
The x coordinate of the eye is
(-ii - 3) = -5
The coordinates of the center are
(-5 , 5)
- In the figure below BC is parallel to DE. The length of side Air conditioning is x and the length of side CE is 4x. What is the area of triangle ADE if the area of ABC is 100 cm 2?
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Solution
Since BC is parallel to DE, the triangles ADE and ABC are similar and the lengths of their corresponding sides related as follows
AB / Advertisement = BC / DE = h / H = AC / AE = ten / 5x = 1/5, where h is the altitude of triangle ABC and H is the altitude of triangle ADE.
The areas A1 and A2 of triangles ABC and ADE are given by
A1 = (i/two)BC * h = 100 , A2 = (one/2)DE * H
Since BC / DE = h / H = one/five
(BC * h) / (DE * H) = 1 / 25
and hence
DE * H = 25 (BC * h)
the area A2 is given past
A2 = (1/2)DE * H = 25 (1/two)(BC * h) = 25 * 100 = 2500 cm ii
- In the figure below is shown a right triangle ABC. What is the volume of the cone obtained past rotating the triangle about the side AB?
Solution.
The cone will have radius AC determined using Pythagora'southward theorem
Air-conditioning = sqrt(102 - 6two) = eight
The volume 5 of the cone of radius r = eight and height h = 6 is given by
5 = (one/three) π r2 h = (one/three) π 64 * 6 = 128π
- If iii(x + y) = 27 and x and y are positive integers, which of these cannot exist the value of x / y?
A) 1/2
B) 8
C) 2
D) one/eight
Eastward) 1/iv
Solution
Let united states of america try to solve the arrangement of equation for each case. We get-go with case A)
A) 3(x + y) = 27 and x / y = 1 / 2
Rewrite system of equation equally
x + y = ix and 10 / y = ane / 2 or 2x = y
Solve to notice 10 and y
10 = three and y = half dozen
We solve case B)
B) three(x + y) = 27 and 10 / y = 8
solutions: x = 8 and y = 1
Nosotros solve case C)
C) 3(10 + y) = 27 and x / y = ii
solutions: ten = 6 and y = three
We solve case D)
D) 3(x + y) = 27 and x / y = 1 / 8
solutions: 10 = i and y = eight
We solve case East)
E) 3(x + y) = 27 and x / y = 1 / 4
solutions: 10 = nine/5 and y = 36/v
In instance E) x and y are not positive integers and is therefore in contradiction. Hence the reply to the question is East)
- [ cos(ten) sin(2x) - 2 sinx ] / [sin2ten cos(10)] =
Solution
Use trigonometric identity sin(2x) = 2 sin(x)cos(ten) to rewrite numerator equally follows
cos(ten) sin(2x) - 2 sinx = cos(x) 2 sin(x)cos(x) - ii sin(x)
= 2sin(ten) (cos(x)2 - 1) = - two sinthree(x)
We now substitute the expression in the numerator past - 2 sin3(ten) and simplify
[ cos(x) sin(2x) - two sinx ] / [sintwoten cos(ten)] = - ii sin3(x) / [sin2x cos(ten)]
= -2 sin(ten) / cos(x) = -2 tan(x)
- Find the smallest value of a positive integer x such that ten 2 + 4 is divisible by 25.
Solution
For x 2 + 4 to exist divisible by 25, information technology must of the form
10 2 + 4 = 25 k , where k is a positive integer greater than 1.
Substitute unlike values of thousand and solve for 10
k = one; x 2 + four = 25 k = 25 , x 2 = 21 , x is not a positive integer.
chiliad = 2; x 2 + 4 = 25 k = 50 , ten 2 = 46 , 10 is non a positive integer.
k = 3; ten 2 + 4 = 25 g = 75 , x 2 = 71 , 10 is not a positive integer.
grand = 4; x 2 + 4 = 25 m = 100 , 10 two = 96 , x is not a positive integer.
grand = five; x 2 + 4 = 25 yard = 125 , 10 two = 121 , ten = 11 and is a positive integer.
The smallest, positive integer, value of ten such that x 2 + iv is divisible by 25 is 11.
- Observe the value of constant a such that ten = 1 is a solution to the equation
a√(x + 3) - 4a | 2x - 1 | = 4
Solution
Substitute x past ane in the given equation and simplify
a√(i + 3) - 4a | 2(1) - 1 | = 4
2a - 4a = 4
Solve for a
a = -2
- g and n are positive integers such that thou > n and m 2n = 46656. Detect m.
Solution
We first write m 2n as (mn)2
thousand 2n = (mdue north)2 = 46656
which gives
kn = √ 46656 = 216
Nosotros at present write 216 as the product of powers of prime numbers
216 = ii*ii*2*3*three*3 = ii3 33 = 6iii
Hence
knorthward = 6iii , where yard = 6 and n = 3 and is smaller than m.
- The root of an equation of the form f(x) = 2 is ten = v. The solution of the equation divers by f(- 2 10 + 1) = two is equal to
Solution
If s exist the solution of f(-2x + 1) = 2, and then f(- 2 s + i) = two
Simply we also know that f(5) = 2 since x = five is a solution to f(x) = 2; hence
f(- two southward + 1) = f(5)
which gives
- two s + 1 = 5
which gives
s = -2
- For what value of b will the system of equations 2x + 5y = 3 and -5x + past = xiv accept no solution?
Solution
The lines corresponding to the two equations given above have different ten-intercepts and therefore the given arrangement of equations will accept no solution if the slopes of the ii lines are equal. We now write the 2 equations in slope intercept grade
2x + 5y = three may be written as y = -(2/v) x + three/v
-5x + by = 14 may be written as y = (v/b) x + fourteen/b
The slopes of the two lines are equal if
-2/5 = 5/b
Solve for b
b = -25 / 2
If b = -25/2, the given system of equation has no solutions.
- Find the real numbers a and b such that three a - b i = (2 - i)(4 + i) where i = √ -1.
Solution
Expand the right paw side of the given equation
iii a - b i = 9 - 2 i
Two complex numbers are equal if their real parts are and their imaginary parts are equal. Hence
3a = ix and -b = -two
Solve to find
a = three and b = ii
- A dealer increased the price of an item past 20%, then increased the toll of the same item by 30%. If x is the original price, what is the price after the two increases?
Solution
Toll after the first increment
x + 20%x = 10 + 0.2 x
Price after the 2d increase
10 + 0.two 10 + 30% (ten + 0.2x) = x + 0.2x + 0.3(x + 0.2x)
= 1.2x + 0.3x + 0.06x = 1.56x
- Two dice are thrown. What is the probability that the sum of the ii numbers obtained is greater than 10?
Solution
When two dice are thrown, there are 36 possible outcomes:(1,1),(1,2),... Three of these outcomes have a sum greater than 10 and these are: (five,6), (half dozen,5) and (six,6). Hence the probability that the sum of the ii numbers obtained is greater than 10 is given by
3 / 36 = 1 / 12
- If x and y are two real numbers such that 3x + 2y = 5 and 5x + 4y = 9, then 4x + 3y =
Solution
Add the right and the left hand sides of the two equations to obtain a new equation.
(3x + 2y) + (5x + 4y) = 5 + 9
Group like terms.
8x + 6y = 14
Divide all terms past to obtain a new equation.
4x + 3y = 7
- The solution prepare of the inequality is given by the interval
A) (5 , + infinity)
B) (-infinity , -1)
C) (-infinity , one) U (5 , +infinity)
D) (-infinity , -1) U (five , +infinity)
E) (-infinity , 0) U (5 , +infinity)
Solution
The graphs of |2x - 4| (chocolate-brown) and that of x + 1 (in bluish) intersect at x = 1 and x = 5. From the graph, information technology can be deducted that |2x - 4| is greater than x + 1 over the intervals
(-infinity , ane) and (5 , + infinity) .
- If g and north are positive integers are such that m / n = 2 / three, which of these values cannot exist values of m and n?
A) g = 12 and n = xviii
B) m = 60 and north = 90
C) m = 34 and n = 51
D) m = 7 and n = 21
E) one thousand = 102 and n = 153
Solution
Fix and reduce the fraction m / n for the given values
A) thousand = 12 and due north = xviii , g / north = 12/eighteen = two/3
B) one thousand = 60 and n = 90 , m / north = 60 / xc = six / nine = 2/iii
C) m = 34 and n = 51 , m / northward = 34 / 51 = 2*17 / 3*17 = iii/2
D) 1000 = 7 and n = 21 , chiliad / n = 7 / 21 = 1/3
m = 7 and n = 21 cannot be the values of m and n such that grand / n = 2/3
- If 35 is the median of the data set including 21, 7, 45, 33, 62 and x, so 10 =
Solution
We first order the known values if the given fix
seven , 21 , 33 , 45 , 62
If 10 = 3 or xiv, the mean volition be equal to
(21 + 33) / 2 = 27
If ten = 33, the mean will exist equal to
(33 + 33) / 2 = 33
If x = 37, the hateful volition exist equal to
(33 + 37) / 2 = 35
So for x = 37, the data set is
7 , 21 , 33 , 37 , 45 , 62 and the mean is 35
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