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financial accounting information for decisions
Questions and Answers of
Financial Accounting Information For Decisions
Variables x and y are related so that, when ln y is plotted on the vertical axis and ln x is plotted on the horizontal axis, a straight-line graph passing though the points (2.5, 7.7) and (3.7, 5.3)
The table shows experimental values of the variables x and y.The variables are known to be related by the equation y = a/x+b, where a and b are constants.a. Draw the graph of lg y against xy.b.
The curve y = xy + x2 – 4 intersects the line y = 3x – 1 at the points A and B. Find the equation of the perpendicular bisector of the line AB.
In triangle ABC, the midpoints of the sides AB, BC and AC are P92, 3), Q(3, 5) and R(-4, 4) respectively. Find the coordinates of A, B and C.
The table shows experimental values of the variables x and y.a. Draw the graph of ln y against x.b. Express y in terms of x.An alternate method for obtaining the relationship between x and y is to
Two points A and B have coordinates (-3, 2) and (9, 8) respectively.i. Find the coordinates of C, the point where the line AB cuts the y-axis.ii. Find the coordinates of D, the mid-point of AB.iii.
Change these angles to radians, in terms of π.a. 10°b. 20°c. 40°d. 50°e. 15°f. 120°g. 135°h. 225°i. 360°j. 720°k. 80°l. 300°m. 9°n. 75°o. 210°
Find, in terms of π, the arc length of a sector ofa. Radius 6 cm and angle π/4b. Radius 5 cm and angle 2π/5c. Radius 10 cm and angle 3π/8d. Radius 18 cm and angle 5π/6.
Find, in terms of π, the area of a sector ofa. Radius 6 cm and angle π/3b. Radius 15 cm and angle 3π/5c. Radius 10 cm and angle 7π/10d. Radius 9 cm and angle 5π/6.
The diagram shows a sector OPQ of a circle with centre O and radius x cm.Angle POQ is 0.8 radians. The point S lies on OQ such that OS = 5 cm.The point R lies on OP such that angle ORS is a right
Change these angles to degrees.a. π/2b. π/6c. π/12d. π/9e. 2π/3f. 4π/5g. 7π/10h. 5π/12i. 3π/20j. 9π/10k. 6π/5l. 3 πm. 7π/4n. 8π/3o. 9π/2
Find the arc length of a sector ofa. Radius 8 cm and angle 1.2 radiansb. radius 2.5 cm and angle 0.8 radians.
Find the area of a sector ofa. Radius 4 cm and angle 1.3 radiansb. Radius 3.8 cm and angle 0.6 radians.
The diagram shows a circle with centre O and a chord AB.The radius of the circle is 12 cm and angle AOB is 1.4 radians.a. Find the perimeter of the shaded region.b. Find the area of the shaded
Write each of these angles in radians correct to 3 sf.a. 32°b. 55°c. 84°d. 123°e. 247°
Find, in radians, the angle of a sector ofa. Radius 4 cm and arc length 5 cmb. Radius 9 cm and arc length 13.5 cm.
Find, in radians, the angle of a sector ofa. Radius 3 cm and area 5 cm2b. Radius 7 cm and area 30 cm2.
The diagram shows a square ABCD of side 16 cm. M is the mid-point of AB.The points E and F are on AD and BC respectively such that AE = BF = 6 cm.EF is an arc of the circle centre M, such that angle
Write each of these angles in degrees correct to 1 decimal place.a. 1.3 radb. 2.5 radc. 1.02 radd. 1.83 rade. 0.58 rad
Find the perimeter of each of these sectors.a.b.c. 1.1 rad 4 cm
POQ is the sector of a circle, centre O, radius 10 cm.The length of arc PQ is 8 cm. Finda. Angle POQ, in radiansb. The area of the sector POQ.
The diagram shows a right-angled triangle ABC and a sector CBDC of a circle with centre C and radius 12 cm. Angle ACB = 1 radian and ACD is a straight-line.a. Show that the length of AB is
Copy and complete the tables, giving your answers in terms of π.a.b. Degrees 45 90 135 180 225 270 315 360 Radians 2n
ABCD is a rectangle with AB = 6 cm and BC = 16 cm.O is the midpoint of BC.OAED is a sector of a circle, centre O. Finda. AOb. Angle AOD, in radiansc. The perimeter of the shaded region. E А D B'
A sector of a circle, radius rcm, has a perimeter of 150 cm. Find an expression, in terms of r, for the area of the sector.
The diagram shows two circles, centres A and B, each of radius 10 cm. The point B lies on the circumference of the circle with centre A. The two circles intersect at the points C and D. The point E
Use your calculator to finda. sin 1.3 radb. tan 0.8 radc. sin 1.2 radd. sin π/2e. cos π/3f. tan π/4.
Finda. The length of arc ABb. The length of chord ABc. The perimeter of the shaded segment. A B 2.3 rad 10 cm 10 cm
ABCD is rectangle with AB = 9 cm and BC = 18 cm.O is the midpoint of BC.OAED is a sector of a circle, centre O. Finda. AOb. Angle AOD, in radiansc. The area of the shaded region. E A B
The diagram shows a circle, centre O, radius 8 cm. The points P and Q lie on the circle. The lines PT and QT are tangents to the circle and angle POQ = 3π/4 radians.i. Find the length of PT.ii. Find
Anna is told the size of angle BAC in degrees and she is then asked to calculate the length of the line BC. She uses her calculator but forgets that her calculator is in radian mode. Luckily she
Triangle EFG is isosceles with EG = FG = 16 cm. GH is an arc of a circle, centre F, with angle HFG = 0.85 radians. Finda. The length of arc GHb. The length of EFc. The perimeter of the shaded region.
The circle has radius 12 cm and centre O.PQ is a tangent to the circle at the point P.QRO is a straight-line. Finda. Angle POQ in radiansb. The area of sector PORc. The area of the shaded region. 35
The diagram shows a circle, centre O, radius r cm. Points A, B and C are such that A and B lie on the circle and the tangents at A and B meet at C. Angle AOB = θ radians.i. Given that the area of
AOB is the sector of a circle, centre O, radius 8 cm.AC is a tangent to the circle at the point A.CBO is a straight-line and the area of sector AOB is 32 cm2.Finda. Angle AOB, in radiansb. The area
Triangle EFG is isosceles with EG = 9 cm.GH is an arc of a circle, centre F, with angleHFG = 0.6 radians, Finda. The area of sector of HFGb. The area of triangle EFGc. The area of the shaded region.
The diagram shows a circle, centre O, radius 12 cm.Angle AOB = θ radians.Arc AB = 9π cm.a. Show that θ = 3π/4.b. Find the area of the shaded region. 9t cm A В 12 cm 12 cm
AOD is a sector of a circle, centre O, radius 4 cm.BOC is a sector of a circle, centre O, radius 10 cm.The shaded region has a perimeter of 18 cm. Finda. Angle AOD, in radiansb. The area of the
AOB is a sector of a circle, centre O, with radius 9 cm.Angle COD = 0.5 radians and angle ODC is a right angle.OC = 5 cm. Finda. ODb. CDc. The perimeter of the shaded regiond. The area of the shaded
FOG is a sector of a circle, centre O, with angleFOG = 1.2 radians.EOH is a sector of a circle, centre O, with radius 5 cm.The shaded region has an area of 71.4 cm2.Find the perimeter of the shaded
The diagram shows a semi-circle, centre O, radius 10 cm.FH is the arc of a circle, centre E.Find the area ofa. Triangle EOFb. Sector FOGc. Sector FEHd. The shaded region. 2 rad E H G
The diagram shows a circle inscribed inside a square of side length 10 cm. A quarter circle of radius 10 cm is drawn with the vertex of the square as its centre. Find the shaded area.
Use the rule logb a = log10 a/log10 b to evaluate these correct to 3 sf.a. log2 10b. log3 33c. log5 8d. log7 0.0025
Use a calculator to evaluate correct to 3 sf.a. e2b. e1.5c. e0.2d. e-3
Convert from exponential from to logarithmic form.a. 103 = 1000b. 102 = 100c. 106 = 1000000d. 10x = 2e. 10x = 15f. 10x = 0.06
Convert from exponential form to logarithmic form.a. 43 = 64b. 25 = 32c. 53 = 125d. 62 = 36e. 2-5 = 1/32f. 3-4 = 1/81g. a2 = bh. xy = 4i. ab = c
At the start of an experiment the number of bacteria was 100.This number increases so that after t minutes the number of bacteria, N, is given by the formula N = 100 × 2t.a. Estimate the number of
Use a graphing software package to plot each of the following family of curves for k = 3, 2, 1, - 1, - 2 and – 3.a. y = ekxb. y = kexc. y = ex + kDescribe the properties of each family of curves.
Sketch the graphs of each of the following exponential functions. [Remember to show the axis crossing points and the asymptotes.]a. y = 2ex – 4b. y = 3ex + 6c. y = 5ex + 2d. y = 2e-x + 6e. y = 3e-x
The following functions are each defined for x ∈ R.Find f-1(x) for each function and state its domain.a. f(x) = ex + 4b. f(x) = ex – 2c. f(x) = 5ex – 1d. f(x) = 3e2x + 1e. f(x) = 5e2x + 3f.
a. Using the substitution y = 5x, show that the equation 52x+1 – 5x+1 + 2 = 2(5x) can be written in the form ay2 + by + 2 = 0, where a and b are constants to be found.b. Hence solve the equation
Write as a single logarithm.a. log2 5 + log2 3b. log3 12 – log3 2c. 3 log5 2 + log5 8d. 2 log7 4 – 3log7 2e. ½ log3 25 + log3 4f. 2 log7(1/4) + log7 9g. 1 + log4 3h. lg 5 – 2i. 3 – log4 10
Solve.a. log2 x + log2 4 = log2 20b. log4 2x – log4 5 = log4 3c. log4 (x – 5) + log4 5 = 2 log4 10d. log3 (x + 3) = 2 log3 4 + log3 5
Solve, giving your answers correct to 3 sf.a. 2x = 70b. 3x = 20c. 5x = 4d. 23x = 150e. 3x+1 = 55f. 22x+1 = 20g. 7x-5 = 40h. 7x = 3x+4i. 5x+1 = 3x+2j. 4x-1 = 5x+1k. 32x+3 = 53x+1l. 34-5x = 2x+4
Given that u = log4x, find, in simplest form in terms of u.a. logx 4,b. logx 16,c. logx 2.d. logx 8.
Use a calculator to evaluate correct to 3 sf.a. ln 4b. ln 2.1c. ln 0.7d. ln 0.39
Solve each of these equations, giving your answers correct to 3 sf.a. 10x = 75b. 10x = 300c. 10x = 720d. 10x = 15.6e. 10x = 0.02f. 10x = 0.005
Convert from logarithmic form to exponential form.a. log2 4 = 2b. log2 64 = 6c. log5 1 = 0d. log3 9 = 2e. log36 6 = ½f. log8 2 = 1/3g. logx 1 = 0h. logx 8 = yi. loga b = c
At the beginning of 2015, the population of a species of animals was estimated at 50,000. This number decreased so that, after a period of n years, the population was 50,000e-0.03n.a. Estimate the
Use a graphing software package to plot each of the following family of curves for k = 3, 2, 1, - 1, - 2 and – 3.a. y = ln kxb. y = k ln xc. y = ln(x + k)Describe the properties of each family of
Sketch the graphs of each of the following logarithmic functions. [ Remember to show the axis crossing points and the asymptotes.]a. y = ln(2x + 4)b. y = ln (3x – 6)c. y = ln (8 – 2x)d. y =
Find f-1(x) for each function.a. f(X) = ln(x + 1), x > - 1b. f(x) = ln(x – 3), x > 3c. f(x) = 2ln(x + 2), x > - 2d. f(x) = 2ln(2x + 1), x > - ½e. f(x) = 3ln (2x – 5), x > 5/2f.
Solve the following simultaneous equations.Log2(x + 3) = 2 + log2 yLog2(x + y) = 3
Write as a single logarithm, then simplify your answer.a. log2 56 – log2 7b. log6 12 + log6 3c. ½ log2 36 – log2 3d. log3 15 – ½ log3 25e. log4 40 – 1/3 log4 125f. ½ log3 16 – 2 log3 6
Solve.a. log6 x + log6 3 = 2b. lg(5x) – lg(x – 4) = 1c. log2(x + 4) = 1 + log2(x – 3)d. log3(2x + 3) = 2 + log3(2x – 5)e. log5(10x + 3) – log5 (2x – 1) = 2f. lg(4x + 5) + 2lg2 = 1 + lg(2x
a. Show that 2x+1 – 2x-1 = 15 can be written as 2(2x) – ½(2x) = 15.b. Hence find the value of 2x.c. Find the value of x.
Given that log9 y = x, express in terms of x.a. logy 9,b. log9 (9y)c. log3y,d. log3(81y).
Without using a calculator find the value ofa. eln 5b. e1/2 ln64c. 3eln2d. -e-ln1/2
Convert from logarithmic form to exponential form.a. lg 1000000 = 5b. lg 10 = 1c. lg 1/1000 = - 3d. lg x = 7/5e. lg x = 1.7f. lg x = - 0.8
Solve.a. log2 x = 4b. log3 x = 2c. log5 x = 4d. log3 x = ½e. logx 144 = 2f. logx 27 = 3g. log2 (x – 1) = 4h. log3 (2x + 1) = 2i. log5 (2 – 3x) = 3
The volume of water in a container, V cm3, at time t minutes, is given by the formulaV = 2000 e-kt.When V = 1000, t = 15.a. Find the value of k.b. Find the value of V when t = 22.
F(x) = e2x + 1 for x ∈ Ra. State the ranger of f(x).b. Find f-1(x)c. State the domain of f-1(x).d. Find f-1f(x)
Functions g and h are defined byg(x) = 4ex – 2 for x ∈ R,h(x) = ln 5x for x > 0.a. Find g-1(x).b. Solve gh(x) = 18.
Simplifya.b. 2log,3 - log, 4 + log,8
Solve.a. log5 (x + 8) + log5 (x + 2) = log5 20xb. log3 x + log3 (x – 2) = log3 15c. 2 log4 x – log4 (x + 3) = 1d. lg x + lg(x + 1) = lg 20e. log3 x + log3(2x – 5) = 1f. 3 + 2 log2 x = log2 (14x
Solve, giving your answers correct to 3 sf.a. 2x+2 – 2x = 4b. 2x+1 – 2x-1 – 8 = 0c. 3x+1 – 8(3x-1) – 5 = 0d. 2x+2 – 2x-3 = 12e. 5x – 5x+2 + 125 = 0
a. Given that logp x = 20 and logp y = 5, find logy x.b. Given that logp X = 15 and logp Y = 6, find the value of logx Y.
Solve.a. eln x = 7b. ln ex = 2.5c. eln x = 36d. e-ln x = 20
Solve each of these equations, giving your answers correct to 3 sf.a. lg x = 5.1b. lg x = 3.16c. lg x = 2.16d. lg x = - 0.3e. lg x = - 1.5f. lg x = - 2.84
Solve.a. logx 64 – logx 4 = 1b. logx 16 + logx 4 = 3c. logx 4 – 2logx 3 = 2d. logx 15 = 2 + logx 5
Without using a calculator, find the value ofa. lg 10,000b. lg 0.01c. lg √10d. lg (3√10)e. lg (10√10)f. lg (1000/√10).
Find the value ofa. log4 16b. log3 81c. log4 64d. log2 0.25e. log3 243f. log2 (8√2)g. log5 (25√5)h. log2 (1/√8)i. log64 8j. log7 (√7/7)k. log5 3√5l. log3 1/√3
A species of fish is introduced to a lake.The population, N, of this species of fish after t weeks is given by the formulaN = 500 e-0.3t.a. Find the initial population of these fish.b. Estimate the
F(x) = ex for x ∈ R. g(x) = ln 5x for x > 0a. Findi. fg(x)ii. gf(x).b. Solve g(x) = 3f-1(x).
Given that logP X = 5 and logP Y = 2y finda. logP X2,b. logP 1/X,c. logXY P.
a. Express 16 and 0.25 as powers of 2.b. Hence, simplify log3 16/log3 0.25.
Use the substitution y = 3x to solve the equation 32x + 2 = 5(3x).
Evaluate logp 2 × log8p.
Solve, giving your answers correct to 3 sf.a. ex = 70b. e2x = 28c. ex+1 = 16d. e2x-1 = 5
Simplify.a. logx x2b. logx 3√xc. logx (x√x)d. logx 1/x2e. logx (1/x2)3f. logx (√x7)g. logx (x/3√x)h. logx (x√x/3√x)
The value, $V, of a house n years after it was built is given by the formulaV = 250,000 ean.When n = 3, V = 350,000.a. Find the initial value of this house.b. Find the value of a.c. Estimate the
F(x) = e3x for x ∈ R. g(x) = ln x for x > 0a. Findi. fg(x)ii. gf(x).b. Solve f(x) = 2g-1(x).
Solve.a. log9 3 + log9(x + 4) = log5 25b. 2 log4 2 + log7(2x + 3) = log3 27
a. Given that log4 x = 1/2, find the value of x.b. Solve 2 log4 y – log4(5y – 12) = ½,
Simplify.a. log7 4 / log7 2b. log7 27 / log7 3c. log3 64 / log3 0.25d. log5 100 / log5 0.01
Solve.a. (log5 x2) – 3log5(x) + 2 = 0b. (log5 x)2 – log5(x2) = 15c. (log5 x)2 – log5(x3) = 18d. 2(log2 x)2 + 5log2 (x2) = 72
Solve, giving your answer correct to 3 sf. za. 32x – 6 × 3x + 5 = 0b. 42x – 6 × 4x – 7 = 0c. 22x – 2x – 20 = 0d. 52x – 2(5x) – 3 = 0
Use the substitution u = 5x to solve the equation 52x – 2(5x+1) + 21 = 0.
Solve, giving your answers in terms of natural logarithms.a. ex = 7b. 2ex + 1 = 7c. e2x-5 = 3d. 1/2 e3x-1 = 4
Solve.a. log3 (log2 x) = 1b. log2 (log5 x) = 2
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