New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'checkley77'
checkley77 answered: 12560 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299 , 12300..12329 , 12330..12359 , 12360..12389 , 12390..12419 , 12420..12449 , 12450..12479 , 12480..12509 , 12510..12539 , 12540..12569, >>Next
Linear_Algebra/234341: You save money from a job. You invest some of it in a cd account that yields 8% interest yearly and the rest in a bond that yields 6% interest yearly
C=the amount of money invested in the cd and B=the amount invested on the bond. You earn $84 interest in a year
Write an equation relating C,B and the amount of money earned
1 solutions
Answer 172791 by checkley77(12569)   on 2009-11-02 19:47:34 (Show Source):
You can put this solution on YOUR website!.08C+.06B=84
WITHOUT ANY ADDITIONAL INFORMATION YOU HAVE ONE EQUATION WITH 2 UNKNOWNS WHICH WILL NOT YIELD A UNIQUE SOLUTION.
THERE ARE ALMOST AN INFINITE NUMBER OF POSSIBLE SOLUTIONS HERE.
EXAMPLES:
LET C=1000 & B=66.67
.08*1000+.06*66.67=84
80+4=84
84=84
OR:
B=500
C=675
675*.08+.06*500=84
54+30=84
84=84
|
Equations/234309: Three sides of a triangle are represented by x,x+3 and x+5. The perimeter
of the triangle is 35 units. Solve for x
1 solutions
Answer 172787 by checkley77(12569) on 2009-11-02 19:28:52 (Show Source):
You can put this solution on YOUR website!x+x+3+x+5=35
3x=35-8
3x=27
x=27/3
x=9 for the short side.
9+3=12 for the middle side.
9+5=14 for the longest side.
Proof:
9+12+14=35
35=35
|
Reduction-of-unit-multipliers/234271: I have a lake with a perimeter of 3,675 ft and I would like to estimate the acerage. I understand that ft. is a mesure of length and acre is a measure of land but I thought there was a way.
thank you
1 solutions
Answer 172778 by checkley77(12569) on 2009-11-02 18:50:56 (Show Source):
You can put this solution on YOUR website!Thank you. I would call it an irregular quadrilateral with the 1st side=429ft, 2nd side=602ft, 3rd side=475 ft, and the 4th side=718ft.
Using these figures and constructing a diagonal I have 2 non-right triangles:
718 by 475 by 801
602 by 429 by 801
Using the formula:
Area=sqrt[s(s-a)(s-b)(s-c)] where s=(a+b+c)/2
s=(718+475+801)/2=1,994/2=997
area=sqrt[997(997-718)(997-475)(997-801)
area=sqrt[997*279*522*196]
area=sqrt[28,459,412,860]
area=168,699 ft^2 for one of the triangles.
s=(602+429+801)/2=1,832/2=916
area=sqrt[916(916-602)(916-429)(916-801)]
area=sqrt[916*314*487*115]
area=sqrt16,108,382,120
area=126,919 ft^2 for the other triangle.
168,699+126,919=295,618 ft^2 estimaate for the area of the lake.
|
Miscellaneous_Word_Problems/234275: Please help me solve this word problem:
Stu has twice as much money invested in stocks as in bonds. Stocks earn 8% interest per year and bonds 2% per year. If Stu earned a total of $400 dollars from his stocks and bonds last year, how much money did he have invested in stocks?
----- Thank you =)
1 solutions
Answer 172776 by checkley77(12569) on 2009-11-02 18:39:57 (Show Source):
You can put this solution on YOUR website!.08*2x+.02x=400
.16x+.02x=400
.18x=400
x=400/.18
x=2,222.22 invested @ 2% in bonds.
2*2,222.22=4,444.44 invested @ 8% in stocks.
Proof:
.08*4,444.44+.02*2,222.22=400
355.55+44.44=400
399.99~400
|
Age_Word_Problems/234276: Please help me solve this word problem:
John's present age is three-fourths of Sally's present age. In five years, John's age will be four-fifths of Sarah's age at that time. What are the present ages of John and Sally?
----- Thank you =)
1 solutions
Answer 172775 by checkley77(12569) on 2009-11-02 18:32:37 (Show Source):
You can put this solution on YOUR website!John's age now=3x/4
Sally's age now=x
In 5 years John will be 3x/4+5
Sally will be 4/5(x+5)
3x/4+5=4/5(x+5)
3x/4+5=(4x+20)/5
(3x+4*5)/4=(4x+20)/5 crosss multiply.
5(3x+20)=4(4x+20)
15x+100=16x+80
15x-16x=80-100
-x=-20
x=20 ans. for sally's age now.
3/4*20=60/4=15 ans. for John's age now.
Proof:
3/4*20+5=4/5(20+5)
15+5=4/5*25
20=100/5
20=20
|
Coordinate-system/234284: Write the slope-intercept form of the equation. (-1,-8) and (-1,9)
1 solutions
Answer 172769 by checkley77(12569) on 2009-11-02 18:16:51 (Show Source):
You can put this solution on YOUR website!(-1,-8) and (-1,9)
Y=mX+b is the line equation when m=slope & b=the y intercept.
m=(y2-y1)/(x2-x1)
m=(9+8)/(-1+1)=17/0 this indicates that the slope is a verticle line through
x=-1 is the line equation.
There is no y intercept.
|
Inequalities/234203: if you have an inequality and you are replacing the variable with a number to see if it is a solution to that given inequality, and the problem looks like this:
4*-2+7>15
do you multiply by negative 2 to both sides?
1 solutions
Answer 172756 by checkley77(12569) on 2009-11-02 18:01:13 (Show Source):
You can put this solution on YOUR website!The proper procedure is to multipy the 4*-2.
4*-2+7>15
-8+7>15
-1>15 This says that you don't have the correct solution.
Unless you meant:
4(-2+7)>15
4*5>15
20>15 Now you have a valid proof.
|
Pythagorean-theorem/234204: a women travels one mile due north, then two miles due east, then three miles due north again, and then once more east for 4 miles. how far is she from her starting point? (it is less than 5+ square root 5 miles.
1 solutions
Answer 172748 by checkley77(12569) on 2009-11-02 17:42:14 (Show Source):
You can put this solution on YOUR website!She is 4 miles north & 6 miles east of her starting location.
Thus you have a right triangle with sides of 4 & 6 miles and need to find the hypotenuse.
a^2+b^2=c^2
4^2+6^2=c^2
16+36=c^2
52=c^2
c=sqrt52
c=7.211 miles is her distance from her starting point.
|
Linear_Algebra/234205: A motorboat can go 14 miles downstream on a river in 20 minutes. It takes 30 minutes for this boat to go back upstream the same 14 miles. Find the speed of the current.
1 solutions
Answer 172746 by checkley77(12569) on 2009-11-02 17:35:43 (Show Source):
You can put this solution on YOUR website!D=RT OR R=D/T
R=14/(1/3)
R=14*3=42 MPH DOWNSTREAM.
R=14/(1/2)
R=14*2/1=28 MPH UPSTREAM.
(42-28)/2=14/2=7 MPH IS THE SPEED OF THE CURRENT.
|
percentage/234206: One sixth of all sales are for cash. Cash sales for the week were $265, what were the total sales.
1 solutions
Answer 172743 by checkley77(12569) on 2009-11-02 17:29:06 (Show Source):
You can put this solution on YOUR website!SORRY I MISTYPEDTHE 21 INSTEAD OF THE 2 & DIDN'T CHECK IT.
I'VE BEEN AT THE COMPUTER FOR ABOUT 5 HOURS AND AM GETTING BUG-EYED.
CORRECTION FOLLOWS.
x/6=265
x=265*6
x=$1,590 ans. for the total sales for the week.
|
Money_Word_Problems/234215: the future value of an investment of $10,000 at 0.4% per month compounded monthly after 17 years
1 solutions
Answer 172739 by checkley77(12569) on 2009-11-02 17:22:02 (Show Source):
You can put this solution on YOUR website!P=(1+r/n)^t*n Where P=principle, r=rate as a decimal, n= thr number if interest periods per year & t=the number of years.
10,000(1+.004/12)^17*12
10,000(1+.003333)^204
10,000(1.003333)^204
10,000*1.971513=$19,715.13 ans.
|
Money_Word_Problems/234216: Calculate, to the nearest cent, the future value of an investment of $10,000 at 6.5%/year, compounded daily (assume 365 days/year) after 9 years
1 solutions
Answer 172737 by checkley77(12569) on 2009-11-02 17:16:01 (Show Source):
You can put this solution on YOUR website!P(1+r/n)^t*n Where P=principle, r=rate as a decimal, n=number of interest periods per year & t=number of years.
10,000(1+.065/365)^9*365
10,000(1+.000178)^3285
10,000(1.000178)^3285
10,000*1.794413=17,944.13 ans.
|
Age_Word_Problems/234217: There are 22 students taking the late bus home. There are 14 more girls than boys on the bus. How many girls and how many boys are taking the late bus home.
1 solutions
Answer 172732 by checkley77(12569) on 2009-11-02 17:09:08 (Show Source):
|
Triangles/234222: a rectangle is 2 feet long and 1.5 feet wide,, what is the distance if you measure it diagonally?
1 solutions
Answer 172730 by checkley77(12569) on 2009-11-02 17:02:09 (Show Source):
You can put this solution on YOUR website!These 2 sides 2 ft. & 1.5 ft. along with the diagonal of the rectangle form a right triangle.
Thus we have the formula for the 3 sides of a right triangle:
a^2+b^2=c^2
2^2+1.5^2=c^2
4+2.25=c^2
c^2=6.25
c=sqrt6.25
c=2.5 ft. ans. for the length of the long side (hypotenuse) of the triangle and the diagonal of the rectangle.
Proof:
2^2+1.5^2=2.5^2
4+2.25=6.25
6.25=6.25
|
Numbers_Word_Problems/233925: If the first and third of three consecutive odd integers are added, the result is 57 less than five times the second integer. Find the third integer.
1 solutions
Answer 172574 by checkley77(12569) on 2009-11-01 20:42:32 (Show Source):
You can put this solution on YOUR website!Let x, x+2 & x+4 be the 3 odd integers.
x+x+4=5(x+2)-57
2x+4=5x+10-57
2x-5x=10-57-4
-3x=-51
x=-51/-3
x=17 ans for the first odd integer.
17+2 is the second osdd integer.
17+4=21 is the third odd integer.
Proof:
17+21=5*19-57
38=95-57
38=38
|
Expressions-with-variables/233834: Please help me solve this problem. I would appreciate it very much. In 2003 an algebra student could have calculated thier estimated final grade (g) for that year using the equation g=.35t+50, in which t represents the students combined test grades. If the student had a combined test grade between 82 and 94, what is the final grade of that student?
1 solutions
Answer 172570 by checkley77(12569) on 2009-11-01 20:27:04 (Show Source):
|
Angles/233859: need to find the measure of an angle whose supplement measures 38 degrees less than three times its complement.
1 solutions
Answer 172561 by checkley77(12569) on 2009-11-01 20:11:11 (Show Source):
You can put this solution on YOUR website!180-X=3(90-X)-38
180-X=270-3X-38
-X+3X=270-180-38
2X=52
X=52/2
X=26 DEGREE IS THE ANGLE.
PROOF:
180-26=3(90-26)-36
154=3*64-38
154=192-38
154=154
|
Pythagorean-theorem/233852: a utility pole breaks and falls. nine feet of the pole remains standing ant the top of the pole lands 12 feet from the base forming a right triangle. find the original height of the utility pole.
1 solutions
Answer 172543 by checkley77(12569) on 2009-11-01 19:49:40 (Show Source):
|
|
