New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'robertb'
robertb answered: 4011 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, >>NextTriangles/456884: The sides of a length of a triangle are x, x+4, and 20, where 20 is the longest side. For which range of values is x an acute triangle
1 solutions
Answer 313550 by robertb(4012)   on 2011-06-02 22:12:03 (Show Source):
You can put this solution on YOUR website!For the triangle to be acute,  , where c is the longest side of the triangle. Note that it is given that the longest side has length 20.
Hence, we must have  , or , after simplifying,  , or (x+16)(x - 12) >0. Since x must have positive values only, x > 12.
From the triangle inequality, we get the relation x +(x+4) > 20, or x > 8. Also, since 20 is the longest side, we must have x + 4 < 20, or x < 16. Hence from the initial conditions, we must have 8 < x < 16.
Intersect the preceding interval with the interval x > 12.
Therefore, for the triangle to be acute, we must have 12 < x < 16.
|
Surface-area/456855: The area of a regualr Octogon is 50 Cm2. How do you find the area of a regular octogon with sides 3 times as large?
1 solutions
Answer 313510 by robertb(4012) on 2011-06-02 20:33:53 (Show Source):
|
Permutations/456128: A poker hand consists of five cards from a standard deck of 52. How many poker hands are there that don't have any spades in them?
1 solutions
Answer 313110 by robertb(4012) on 2011-06-01 02:23:44 (Show Source):
|
Probability-and-statistics/456123: a set of 12 cards is numbered 1,2,3...,12. suppose you pick a card at random without looking . find the probability of each event.
P (a multiple of 3)
P(a multiple of 4)
P(an even number)
1 solutions
Answer 313109 by robertb(4012) on 2011-06-01 02:17:25 (Show Source):
|
Money_Word_Problems/455441: Find the equation of the circle with C(-4,2) an is the tangent to the linne 3x+4y-16=0
1 solutions
Answer 313107 by robertb(4012) on 2011-06-01 01:48:41 (Show Source):
You can put this solution on YOUR website!The distance of the center (-4,2) from the line 3x + 4y - 16 = 0 can be found from the formula  . This distance becomes the radius of the circle. Hence the radius of the circle is
 .
Therefore the equation of the circle is
 .
|
Evaluation_Word_Problems/455359: The time required to finish a test is normally distributed
with a mean of 60 minutes and a standard deviation of 10 minutes. What is the z-Score for a student who finishes the test in 45 minutes?
1 solutions
Answer 313105 by robertb(4012) on 2011-06-01 01:37:00 (Show Source):
|
expressions/456138: Obtain the general solution
dr = b(cosx dr + rsinx dx)
ANSWER: r= c(1 - bcosx)
1 solutions
Answer 313104 by robertb(4012) on 2011-06-01 01:32:38 (Show Source):
You can put this solution on YOUR website!dr = b(cosx dr + rsinx dx)
<==> 
<==> 
<==> 
==> ln r = ln(1 - bcosx) + k, k an arbitrary constant.
<==> r = C(1 - bcosx), after raising e to both sides of the previous equation. (C an arbitrary constant.)
|
Functions/456140: Will you assist me in answering this question. State the degree of the following polynomial 9x^4y^3+6xy-3y.
Iam not sure where to begin or how to answer. Thank you
1 solutions
Answer 313102 by robertb(4012) on 2011-06-01 01:21:05 (Show Source):
|
expressions/456139: Obtain the general solution
y lnx lny dx + dy = 0
1 solutions
Answer 313101 by robertb(4012) on 2011-06-01 01:18:51 (Show Source):
You can put this solution on YOUR website!Divide both sides by ylny, to get
 .
<==> 
Integrate both sides:
 k an arbitrary constant.
After integration by parts (on the first integral), the equation becomes
xlnx - x + ln(lny) = k.
Simplifying:
<==> 
<==> 
<==>  <==>  , where C is an arbitrary constant.
|
Linear_Algebra/455730: Let v1 and v2 be non zero vectors in a vector space V. Show that the following statements are equivalent.
a) v1 is not in Rv2=span{v2}
b)Rv1 doesn't equals Rv2
1 solutions
Answer 312878 by robertb(4012) on 2011-05-31 00:58:15 (Show Source):
You can put this solution on YOUR website!(a)==>(b):  not in  ==>  for any non-zero real number k.
==> For any non-zero real number m,  , and hence no element of span{v1} can be found in span{v2}.
(b) ==> (a):  ==> there is element  in  that is not in , or ,
 for any non-zero real number r, or
 . Hence v1 is not in span{v2}.
|
Linear_Algebra/455735: If V={v1,v2,...,vk} is linearly independent, and w is not element of V. Then show that {v1+w,v2+w,...,vk+w} is linearly independent
1 solutions
Answer 312877 by robertb(4012) on 2011-05-31 00:38:54 (Show Source):
You can put this solution on YOUR website!Let  +...+  .
Now let
 +...+  =  ,
which for the purpose of contradiction, we assume that not all the a's are 0, (i.e., the set is linearly DEPENDENT).
==> (  +...+  )*
+(  +...+ )*  + ...+(  +...+  )* =
By the hypothesis, {v1, v2, v3,..., vk } is linearly independent, and thus,
+...+ 
+...+
...................................
+...+  .
We get a homogeneous system of equations.
Adding all corresponding sides of the system, we get
 +...+ 
OR,
+...+ .
Subtracting this equation from each one of the equations in the system above, we obtain
 =...= ,
CONTRARY to the initial assumption that not all a's are 0.
Hence {v1+w,v2+w,...,vk+w} must be linearly independent.
|
Complex_Numbers/455727: Compute the special products and write your answer in a+bi form.
(-2+5i)^2
I know I should foil this but I am doing something wrong. Could you help me here showing all steps please.
David
1 solutions
Answer 312872 by robertb(4012) on 2011-05-31 00:09:59 (Show Source):
|
|
