Answers by visitors named: Henry

C=6.28*4.5 = 28.3m/rev.=Circumference. a. V=5.35rev/min * 28.3m/rev=151m/min. c. a = (Vf - Vo) / t, a = (5.35 - 0)rev/min / (29/60)min = 11.1m/min^2.

10b^6 /(3r^2*9r^7/100b), 10b^6/(27r^9/100b), Invert fraction and multiply: 10b^6 * 100b/27r^9, 1000b^7 / 27r^9.

a. 64o/2.70min * 1rev/360o * 1min/60s= = 1.1 rev/s. b. C = pi*D = 3.14*5.37=16.9 Ft=5,12 m. The Circumference. b. 64o/2.70min * 1rev/360o * 5.12m/rev * 1min/60s = 5.62 m/s.

See Related Questions: tue,1-15-13,11:38am.

I = P*r1*t+P*r2*t = $380. I = P1*.05*1+P2*.09*1 = 380 0.05P1 + 0.09P2 = 380. Multiply both sides by 100: Eq1: 5P1 + 9P2 = 38000. Eq2: P1 + P2 = 6000. Multiply Eq2 by -5: 5P1 + 9P2 = 38000 -5P1 + -5P2 = -30000 Add the 2 Eqs: 4P2 = 8000 P2 = $2000 = Investment @ 9%. In Eq2, substitute 2000 for P2: P1 + 2000 = 6000 P1 = $4000 = Investment @ 5%.

b. d1 = 85t1 = 16 km, 85t1 = 16, t1 = 16 / 85 = 0.1882 h = 11.29 min. d2 = 115t2 = 16 km, 115t2 = 16, t2 = 16 / 115 = 0.139 h = 8.35 min. t1 - t2 = 11.29 - 8.35 = 2.94 min. Car #2 arrives 2.94 min. sooner. c. 15 min = 1/4 h = 0.25 h. d1 = d2, 115t = 85(t + 0.25), 115t = 85t + 21.25, 115t - 85t = 21.25, 30t = 21.25, t = 21.25 / 30 = 0.71 h, d = 115 * 0.71 = 81.65 km.

V^2 = Vo^2 + 2g.h. V^2 = 0 + 19.6*8 = 156.8 V = 12.5 m/s.

X = -20 m. Y b= -20 m. D^2 = X^2 + Y^2. D^2 = 20^2 + 20^2 = 800 D = 28.28 m. = Displacement. tanAr = Y/X = -20/-20 = 1. Ar = 45o = Reference angle. A = 180+45 = 225o.

See 9:08am post.

L = V/F. V = L*F = 15*24 = 360 m/s. Note: L = Wavelength.

(1.4^X3+4.8^X2-3.3X)-(-4.9^3-4.3^2+86 RULE FOR SUBTRACTION:change the sign of number being subtracted and add. (1.4X^3+4.8^2-3.3x)+(4.9X^3+4.3X^2-86= 6.3X^3+9.1X^2+3.3X-86.

d1 = Vo*(t+1) + 0.5at^2, d1 = 33(t+1) + 0.5*0*(t+1)^2, d1 = 33(t+1) + 0 = 33t + 33. d2 = 0 + 0.5*2.62*t^2, d2 = 1.31t^2. d1 = d2, 33t + 33 = 1.31t^2, 1.31t^2 - 33t - 33 = 0, Solve using Quadratic Formula and get: t = 26.15s, and -0.96s. Use the positive valuie: t = 26.15s.

Multiply 2 by each term in the parenthesis: 12. 2(X+5) = 2*X + 2*5 = 2X + 10. 13. (X+4)(X-2) = 0. X+4 = 0, X = -4. (X-2) = 0, X = 2. Solution: X = -4, OR X = 2.

We have 2 fractions with a common denominator 0f 3x. 14/(3X) + (X+5)/(3X)=(14 + (X+5)) / 3X = (14 + X + 5) / 3X = (X+19) / 3X.

1. Draw a hor. line from the origin on the positive x-axis(due East). Label it 80km. 2. Draw a ver. line upward from the end of the hor line parallel to the positive y-axis(due North) and label it 192km. If you drove along the given parh, you would have traveled 272km(80km+192km). If you took a st. line from the origin to the final destination, you would have traveled: d = sqrt((80)^2+(192)) = 208km. tanA = Y/x = 192 / 80 = 2.4, A = 67.38 deg. R = X / cosA = 80 / cos67.38 = 208km..

33km @ 90+10 = 33km @ 100 deg.,CCW. 18km @ 270+45 = 18km @ 315 deg.,CCw. 14km @ 270+50 = 14km @ 320 deg.,CCW. X=hor.=33cos100 + 18cos315 + 14cos320 = 5.73 + 12.73 + 10.72 = 17.7km. Y=ver.=33sin100 + 18sin315 + 14sin320 = 2.95 + (-12.73) + 9.0 = -0.78km. TanA = Y / X = -0.78 / 17.7 = -0.04407 A = -2.52 deg = 2.52 deg South of East. D = -0.78 / sin(-2.52) = 17.67km @ 2.52 deg South of East. The return path is 17.67km @ 2.52 deg North of East.

Oh, Thanks sir.

Thank-you also!!

x^3 -4x^2 -x + 4. The expression has 4 terms. Therefore, we can form 2 groups with 2 factorable terms in each group: (x^3 - x) + (-4x^2 + 4), Factor each pair: x(x^2 - 1) -4(x^2 - 1), Factor out (x^2 - 1): (x^2 - 1)(x - 4), (x + 1)(x - 1)(x - 4). NOTE: (X^2 - 1) = (x + 1)(x - 1)

OOPs! 2. u*Fv = Fk 1o6.8u = 66.8 u = 66.8 / 106.8 = 0.625. NOTE: Fv = 106.8 N.

Correction: -Fk = 2.6 - 69.4 = -66.8Fk = 66.8 N. 2. u*Fv = Fk 127.4u = 66.8 u = 66.8 / 127.4 = 0.524 = uk.

Wb = m*g = 13kg * 9.8N/kg = 127.4 N. = Wt. of box. Fb = 127.4N @ 33 Deg. = Foce of box. Fp = 127.4*sin33 = 69.4 N. = Force parallel to incline. Fv = 127.4*cos33 = 106.8 N. = Force perpendicular to incline. 1. Fn = Fp - Fk = ma 69.4 - Fk = 13 * 0.20 = 2.6 -Fk = 2.6 -69.4 = -72 Fk = 72 N = Force of kinetic friction. 2. u*Fv = Fk 127.4u = 72 u = 72 / 127.4 = 0.565 = Coefficient of kinetic friction.

KM = 2LM, 5x - 6 = 2(2x+4), 5x - 6 = 4x + 8, 5x - 4x = 8 + 6, X = 14.

% Financed = 100% - 18% = 82%. Amt. Financed = $24,000 * 0.82 = $19,680. Pt = Po*r*t / (1 - (1+r)^-t). r = 4.5% / 12 = 0.375% = 0.00375 = Monthly % rate expressed as a decimal. Pt=(19680*0.00375*36)/(1-(1.00375)^-36, Pt=2656.80 / 0.126063452 = $21,075.10. Int.=Pt - Po=21075.10 - 19680=$1395.10 Monthly Payments=21,075.10 / 36 = $585.42.

See previous post.

1. 7X + 3Y = 42. Convert the std form Eq to the slope-interceptform: 7X + 3Y = 42, 3Y = -7X + 42, Y = -7/3X + 14, (Y = mx + b). Our 1st point will be at the Y-int where X is always 0. P1(0 , 14). Since our slope is neg. DECREASE b by 7: y = 14 - 7 = 7. X = 3 = The denominator of the slope. P2(3 , 7). GRAPH: P1(0 , 14) , P2(3 , 7) 2. 7X + 3Y = 42 At the y-int., X = 0. Calculate Y: 7(0) + 3Y = 42 3Y = 42 Y = 14. P1(0 , 14). At the X-INT, Y = 0. Calculate X: 7X + 3(0) = 42, 7X = 42, X = 6. P2(6 , 0). GRAPH: P1(0 , 14) , P2(6 , 0).

V(h,k) = V(1,4), P(0,3). Y = a(x - h)^2 + k. P(0,3). Y = a(0 - 1)^2 + 4 = 3, a + 4 = 3, a = -1. Eq: Y = -1(x - 1)^2 + 4. Vertex Form. Y = -x^2 + 2x - 1 + 4. Y = -x^2 + 2x + 3. Solve using Quadratic Formula: X = -1, and 3. Or (-1,0), and(3,0) Input = X. Output = Y. (x,y) (-3,-12) (-2,-5) (2,3) (4,-5)

Does that mean the answer would be the square root of 39419?

Okay thanks.

Y = Elevation. Y =< 1458 meters.

Use 2(3x - 7) + 4 (3 x + 2) = 6 (5 x + 9 ) + 3 to solve

(-3/(X^2-2X+1))+(2X/(X^2-5X+4) In order to find an LCD, we need 2 or more fractions separated by a plus or minus sign. so I added a plus sign. CD=(X^2-2X+1)(X^2-5X+4)=(X-1)^2(X-1)(X-4).This product is a CD but not the LCD. To get LCD, reduce the squared factor to (X-1), LCD=(X-1)(X-1)(X-4)= (X-1)^2(X-4).

Vo = 18m/s @ 52 deg. Vo(hor) = 18cos52 = 11.08m/s. Vo(ver) = 18sin52 = 14.18m/2,up. Vf^2 = Vo^2 + 2gd = 0 @ max height. (14.18)^2 + 2(-9.8)d = 0, 201.07 -19.6d = 0, d = 10.3m. = max barrier that can be cleared.

V=NAh=6(14x12.1/2)10=5082cubic cm V=volume; N=number of sides=number of triangles that forms base of hexagon; A=area of each triangle(quanity in parenthesis); h=height of hexagon. The 6 triangles formed by drawing a line from the center to each vertex are equalateral. Therefore, the height of each triangle is 14sin60=12.1. I hope this explanation is sufficient.

If an equilateral triangle has a perimeter of 18,the length of each side of the triangle is six(18/3=6). Therefore, the perimeter of the hexagon is thirty-six(6x6=36). If a straight line is drawn from the center to each vertex, six congruent equilateral triangles are formed. The perimeter of the hexagon is equal to the sum of the six sides(36).

Factor:X^2 + 14X + 49. 49 = 1*49 = 7*7 Try 1 and 49 in your factored equation: ( X + 1 )*(X + 49). The product of the two binomials does not equal the original equation, Therefore, 1 and 49 is incorrect. Try 7 and 7:(X+7)*(X+7)= X^2 + 14X +49. Yes!

Problem:(X+ )^2. The results will be a perfect square regardless of what number you put in the blank, because you remove the parenthesis by squaring the contents.(X+2)^2=X^2+4X+4.(X+3)^2=X^2+6X+9.Both trinominals are perfect squares.

Mathematical solutions to systems of equations are more accurate than graphical solutions; because when the point of interception fall between grid lines, you must estimate the point location. Which is easier the elimination method or the substitution method? Well, it depends on the equations. Most of the time I use the elimination method. But sometimes I find it easier to use the substitution method. If one of the equations has only one unknown, I use the substitution method. Most of the time it will depend on the individual.

Problem: (X^2+4X-5)/(5X-5)*5X/(X+5) Solution:(X-1)*(X+5)/5(X-1)*5X/(X+5)=(X+5)/25X/(X+5)=(X+5)*(X+5)/25X=(X+5)^2/25X. The parenthesis are neccessary for clarity. Notice that the numerator and a portion of the denominator are factored.

1. C=25P+150. C=Total cost. P=The number of people attending. 2. Slope=25 3. The slope represents the cost per person.

2. C1=C2 7.95+.05X=9.95+.03X .05X-.03X=9.95-7.95 X=100 Minutes CI=Cost at 1st company C2=Cost at 2nd company

15. Rate upstream=30/4=7.5m/hr Rate downstream=30/2.5=12m/hr Rate in stillwater=(7.5+12)/2=9.75m/hr

20. 5X^2-10X=0 5X(X-2)=0 5X=0, X=0 X-2=0, X=2 Solution set:x=0,and x=2.

50. (x-1)(x^3+x^2+x)=x^4+x^3+x^2-x^3-x^2-x=x^4-x=x(x^3-1)

P+Int.=1000000. P=Principal=Total Dep. P+(0.03)(30)P=1000000 P+0.9P=1000000. P=526315.79=Total dep./30years Dep./yr=526315.79/30=17543.81

2X-Y=1. (0,-1), (1,1) 8X-4Y=8. (0,-2), (1,0) The solution of the two equations is the point where their graphs intersect. The graph using the above points showed that the lines are parallel and do not intersect. Therefore. there is no solution. The two lines have equal slopes which further proves that they are parallel. Slope1=-A/B=-2/-1=2, Slope2=-8/-4=2. If you solved the two equations mathematically using the elimination method.the results will be 0=4 which is not a true statement and further proves that there is no solution.

Sun/Sat.=1.988x10^27/5.688x10^23= 0.3495x10^4=3.495x10^3

(8a^2.b^5)(-7a.3b)=-168a^3.b^6 Hint: Multiply the constants first.

3^31=6.1767X10^14=0.61767X10^15 9X3^30=1.8530X10^15 3^33=5.5591X10^15=The greater number Hint: All 3 numbers were raised to the same exponent for ease of comparison. For instance,5.56 is greater than 0.62 and 1.85.

Factor 6X-X^3-X^2 completely. Step1. Arrange exponents in decreasing order: -x^3-x^2+6x. Step2.-X(X^2+X-6) Step3. -X(X-2)(X+3). The equation in step2 is not completely factored, because the trinomial can be factored into 2 binomials (step 3).

4. 27Y^4-147Y^2=3Y^2(9Y^2-49)= 3Y^2(3Y+7)(3Y-7) Hint: 9y^2-49 is the difference of 2 squares and is factored into 2 binomials.

F(x)=5X+1/4 Or Y=5X+1/4 To determine the inverse of a function, replace all Xs with Y and all Ys with X: X=5Y+1/4 Then Multiply both sides of eguation by 4 to eliminate the fraction: 4X=20Y+1. Solve for Y: Y=(4X-1)/20= Inverse function.

14a. Y=mX+b, m=slope, b=Y-intercept. Equation:Y=4/3X-2 14b. Given: Y=-2X+5, p(-2,6). m1=-2 m2=the negative reciprocal of m1: m2=1/2, Y=mx+b, 6=1/2(-2)+b, b=7, Equation: Y=1/2X+7.

If the Central Angle is 98 degrees, the arc is also 98 degrees. Since the arc is less than 180 degrees, it is a minor arc.

D^2=(5/3-2/3)^2+(7-4)^2 D^2=1+9=10 D=sqrt10=3.16

(X-h)^2+(y-k)^2=r^2. h=x-coordinate of the center, k=y-coordinate of the center. (x-2)^2+(y-0)^2=5=r^2. C(h,K)=(2,0), r^2=5, r=sqrt5=2.24

Given:X^2=5X+2. Step1. X^2-5X=2 Step2. X^2-5X+(5/2)^2=2+(5/2)^2 Step3. X^2-5X+25/4=2+25/4. Step4. (X-5/2)^2=33/4 Step5. Take sqrt of both sides. X-5/2=+/-sqrt33/2 Step6. X=5/2+/-sqrt33/2. Step7. Place equation in step6 over a common denoninator. X=(5+/-sqrt33)/2=5.37 and 0.37

The line segment is a vertical line located on the y-axis, and its' mid- point is P(0,-1.5). Your answer is CORRECT!!

Given:CosA=SQRT2/2 Since the cosine is positive and the tangent and cotangent are negative (less than 0), we are in the 4th Quad- rant. CosA=sqrt2/2=x/r, x=sqrt2, r=2, Y=sqrt(r^2-x^2)=sqrt(4-2)=+/-sqrt2, Y=-sqrt2(4th Quadrant). SinA=y/r=-sqrt2/2. CosA=sqrt2/2=x/r(Given) TanA=y/x=-sqrt2/sqrt2=-1 cscA=-(2/sqrt2). secA=2/sqrt2 CotanA=sqrt2/-sqrt2=-1

Given: X=2Y^2+3 Vertex form:X=(y-k)^2+h, h and k are the x and y-coordinates of the vertex. X=2(y-0)^2+3, V(h,k)=(3,0), Directrix: X=h-1/4a=3-1/8=2 7/8,Y=K=0, D(2 7/8,0). Focal: X=h+1/4a=3+1/8=3 1/8, Y=k=0, F(3 1/8,0). Axis: Y=0 (A horizontal line on the X-axis).

9.32ml/sec=.00932L/sec=.00932/(1/60)L/min=0.5592L/min=0.5592L/(1/60)hr=33.55L/hr

57.3qtsx32 oz/qt=1833.6 oz. 1833.6 ozx29.6mL/oz=54274.56mL 54274.56mL/25mL=2171 Vials

V=3.14X1.5^2X5.5/3=13.0cubic inches.

V=(3.14X1.5^2X5.5)/3=13.0cubic inches.

12.2FT/SEC=0.00231MILES/SEC =0.00231MILES/(1/60)min =0.1386miles/1min =0.1386 MILES/(1/60)hr =8.316miles/hr 8.316 miles /hr X 24 hrs/da=199.6 miles/da 25000 miles/199.6 miles/da=125 days.

1 meter=3.3ft, 56200 meters=185460ft. 56.2km/s=185460ft/s=35.125 miles/s =35.125miles/(1/60)min =2107.5 miles/min

1 yr=365.25 days. 1x10^9yrs = 365.25 x10^9 days. 1 day = 24 hrs. 365.25 x10^9 days = 24 x 365.25 x 10^9 = 8766 x 10^9 hours. 1 hr = 60 min. 8766 x 10^9 hrs = 60 x 8766x10^9 hrs= 525960 x 10^9 min =5.25960 x 10^14 min

1. 3X-4<10-4X. 2(X-5)>3(2X-6) 3X-4<10-4X, 7X<14, X<2. 2(X-5)>3(2X-6), 2X-10>6X-18, -4X>-8, X<2. Since the solution to both inequalities is the same(X<2),they are equivalent.

2. Cost=2.49X + .6X +3.75=3.09X +3.75 X=Number of jars purchased. 3.75=Handling charges=constant. 4. GIVEN: (-2,3), (-4,6) Slope=m=(Y2-Y1)/(X2-X1)=(3-6)/(-2-(-4))=-3/2, Y=mX+b,3=-3/2(-2)+b,b=0,Y=-3/2X b=Y-intercept.

SQUARE: A=S^2=S.S, Perimeter=P=4S. REGULAR HEXAGON: Perimeter=T=6h,h=T/6. h+P=T/6+4S.

DIAGONALS: 12 and 16 cm. The diagonals of a rhombus are perpendicular bisectors of each other and form 4 congruent right triangles. The legs of a triangle are: X=12cm/2=6cm, and Y=16cm/2=8 cm. The length of each side is:S^2=X^2+Y^2= 6^2+8^2=100sq.cm, S=10cm. NOTE:All sides of the rhombus are equal.

If the problem means 4 times the sqrt (81/16),then Reiny's answer(9) and procedure is correct. If the problem is asking for the 4th root of(81/16), then the answer is a(3/2).

The Mean Deviation is the average of the five numbers: (2+4+6+5+3)/5=4

Yes, you can solve the exponential equation without using logs if you are using a calculator. EQUATION:L=2.43*W^0.3326=2.43*24^0.3326=2.43*2.8778=6.99=7.0 In this example, L was the unknown and W was 24lbs.The calculator was used to solve expo. If W was the unknown and L was 7, the equation would be: 7=2.43*W^0.3326, Divide both sides by 2.43: W^0.3326=2.8764,W=23.97=24.0 You will have to KNOW your calculator.

sqrt(3)b+sqrt(2)=sqrt(7)b-10 sqrt(3)b-sqrt(7)b=-10-sqrt(2) 1.732b-2.646b=-11.414 -0.914b=-11.414 b=12.49 Your answer(a) is correct!

(X+5)/(X-3). The non-permissible value of X is the value that makes the denominator =0: X-3=0, X=3(non- permissible).

3sqrt(-27)=3sqrt(9*3*-1)=3*3i*sqrt3= 9i*sqrt3. imagenary solution: d

This inequality means that Z can vary from -1.1 to 0. Therefore, Z=-1.1min. Z=0,max.

Pt=Po*(r+1)^n. Compounded daily. Pt=principal at maturity. Po=Initial principal or deposit. r=DPR=Daily percentage rate. n=The number of compounding periods. t=6mo=0.5yr.=time for maturity. Po=1500. APR=4.0 r=4/365/100=0.000109589 n=0.5*365=182.5 days. Pt=1500*(0.000109589+1)^182.5=1530.30

Logb^3*sqrtX^8/(Y^2)(Z^5)= Logb^3*sqrtX^8-Log(Y^2)(Z^5)= Logb^3+LogsqrtX^8-LogY^2+LogZ^5.

Kathy travels 57km/h faster and 21 miles farther(57-36) in the same period of time. Therefore r*t=d, 57*t=21. t=21/57=0.3684 hours for each rider. r*t=d, r*0.3684=36, r=97.7km/h for Carlos. r*0.3684=57, r=154.7km/h for Kathy.

LogmP=a, m^a=P.

RATIO=(86.25-52.75)/86.25=33.5/86.25= 6.7/17.25

Original amount=X Dollars. Bought 50 dollar radio. Bal.=X-50 (X-50)-3/5*(X-50)=2/5*(X-50) Bal. 1/2*2/5*(X-50)=1/5*(X-50) Bal. 1/5*(X-50)=35. Solve for X: X=225=Original Amt.

a. Option A.1575000 *120mo.=189000000 in 10 yrs. b. Option B.1.575M @ 18% APR,Compounded monthly. Pt=Po*(r+1)^n. Pt=Value at 10 yrs. r=MPR=Monthly percentage rate. n=the number of interest compounding periods. r=18/12/100=0.015 n=12*10=120 Pt=1575000*(0.015+1)^120=9401683.52= Value @ 10yrs. Evidently, this is not a practical situation.

The ladder forms a rt. triangle with the wall and ground. The wall is the height.When the height is 10ft, X^2+10^2=16^2, X=12.49ft. When h goes to zero(ladder flat on ground), X^2+0^2=16^2, X=16, t=10ft/5ft/s=2sec. dx/dt=(16-12.49)ft/2s=1.75ft/s

LAW OF COSINES: CosA=(b^2+c^2-a^2)/2bc Cos46=(16+64-a^2)/2*4*8. Cos46=(80-a^2)/64, 80-a^2=64Cos46, a^2=80-64Cos46=35.54, a=5.96=6.0 LAW of SINES: SinB/b=SinA/a SinB/4=Sin46/6,Multiply both sides by 4: SinB=4*Sin46/6=0.4791, B=28.66=28.7

(8r+16)/(24r-24)*(6r-6)/3r+6), FACTOR: 8(r+2)/24(r-1)*6(r-1)/3(r+2), 8/24*6/3=1/3*2/1=2/3

The non-permissable values results in a denominator of 0.Let X^2+X-12=0, Solve for X by factoring:(X+4)*(X-3)=0 X+4=0, X=-4; X-3=0, X=3. Ans.= d

L/P=L/(2W+2L)=1/3. Cross multiply and solve for W: 2W+2L=3L, 2W=3L-2L, 2W=L, W=L/2. CONCLUSION:The width is 1/2 of the Length. Sketch a rectangle and L=4, W=2. L/P SHOULD EQUAL1/3. CHECK:L/(2W+2L)= 4/(4+8)=1/3.

Yes, your answer is correct!

The altitude of an isosceles triangle is the perpindicular bisector of the base and devides the triangle into 2 congruent rt. triangles. Y=12=altitude, X=18/2=9= 1/2 of base. TanB=Y/X=12/9=1.3333, B=53.13 Degrees.

X=3, X-3=0. X=4-i, x-4=-i, Square both sides:X^2-8X+16=-1,X^2-8X+17=0, Multiply:(X-3)(X^2-8X+17)=0 X^3-11X^2+41X-51=0. Yes, the correct choice is C.

1. Correct. 2. Correct. 3.Incorrect. If the pie was cut into 16 slices,each slice is 1/16th of the total area.A 1/2 slice would be 1/32nd (1/2*1/16)of the total area: A=3.14*10^2/32=9.8sq.in. 4.You can cut the pizza into as many piecse as you like; the area will be the same if you do not change the radius. A=3.14*10^2=314 sq.in. 5. If cut properly the radius will be the same, but the area will be 3.14*10^2/32. I hope this will help.

Since the 2 angles are equal, the 2 sides are also equal.We'll call the horizontal side X and the vertical side Y. X=12*Cos45=8.48, Y=12*Sin45=8.48. NOTE: the sin45=Cos45.

It is 1/2 of an Equilateral triangle.

Because the two interest rates are slightly different,the investment at 8% will be less than 2/3 of the 7.2% investment by an amount equal to the difference between the interest rates. 2/3=0.6666, 0.6666-(.08-.072)= 0.6586. The investment at 7.2%=X dollars, The investment at 8%=0.6586X dollars. X+0.6586X=24000, X=14470.03 0.6586X=9529.96 INV @ 7.2%=14,470.03 INV @ 8%=9529.96

The ladder and tree form a right triangle with the ground, The height (h) is the distance from the base of the tree to the point where the ladder touches the tree. Sin65=h/X=h/5, h=5Tan65=10.7 Feet

All of the angles of an equilateral triangle are equal: A=B=C=(7X-3)=60deg. X=9(no units). All of the sides are alsoequal:AB=BC=AC =(3X+12)mm, P=3(3X+12)==9X+36=9(9)+36= 117mm.