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richard1234 answered: 5385 problems
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Geometry_proofs/567032: how can you write an indirect proof that two obtuse angles cannot form a linear pair? 1 solutions
Answer 366462 by richard1234(5390) on 2012-02-02 23:02:43 (Show Source):
You can put this solution on YOUR website!Assume that they do form a linear pair. Since both angles' measures are greater than 90 deg, then their sum must be greater than 180 deg. However, angles form a linear pair if and only if their sum is 180 deg, so contradiction.
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Geometry_proofs/567088: Proof the given is angel one and angel two are right angels prove that angel one is congruent
To angel two in no less than four steps 1 solutions
Answer 366460 by richard1234(5390) on 2012-02-02 22:57:12 (Show Source):
You can put this solution on YOUR website!If you want answers for questions about angels, consult your local church or clergy or read the Bible or something. You want "angles" instead.
If angles 1 and 2 are right angles, then they must have the same degree (or radian) measure, therefore they are congruent.
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absolute-value/567162: If you have a problem like |x-2| + 3 = 5 you would solve it with a positive and negative equation changing the signs of both the 5 and the 3. If you had an inequality |x-2| + 3 > 5 your positive and negative equations would not include changing the sign of the 3. Why is that?
|x-2| + 3 = 5
x + 1 = 5
x=4
|x-2| - 3 = -5
x - 5 = -5
x=0
|x-2| + 3 > 5
x + 1 > 5
x>4
|x-2| + 3 < -5
x + 1 < -5
x<-6 1 solutions
Answer 366455 by richard1234(5390) on 2012-02-02 22:48:20 (Show Source):
You can put this solution on YOUR website!I don't know, your solution is a bit flawed. You can't really say that |x-2| + 3 is always equal to x+1, since there are negative solutions involved too.
You should move the 3 to the RHS first:
Now we can take positive and negative solutions:
 or  , this yields x=4 and x=0. Here, you would change the sign of the 3, but I wouldn't do that; you're more likely to make a mistake this way.
For the inequality, we have
 , this can either be  or  (since we can have a negative solution -x-2 > 2, multiplying by -1 reverses the direction of the inequality). The solutions are x > 4 and x < 0.
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Exponents/567073: im knew to algebra, and i don't know how to attack it, do u think u can help me out.
1. If the average of three numbers is V. If one of the numbers is Z and another is Y, what is the remaining number?
A. ZY - V
B. Z/V - 3 - Y
C. Z/3 - V - Y
D. 3V- Z - Y
E. V- Z - Y
1 solutions
Answer 366452 by richard1234(5390) on 2012-02-02 22:41:29 (Show Source):
You can put this solution on YOUR website!Let X be the remaining number. Since the average of X,Y,Z is V, we have
Subtract Y+Z from both sides to isolate X.
This is the same as answer choice D.
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Probability-and-statistics/567102: A student is to select 3 classes for next semester. If this student decides to randomly select one course from each of 5 economics classes, 9 math classes, and 5 conputer classes, how many sifferent outcomes are possible. Please explain. 1 solutions
Answer 366451 by richard1234(5390) on 2012-02-02 22:39:37 (Show Source):
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Geometry_proofs/567156: How would you solve this? I'm in ninth grade in Geometry 1 Honors: "Find the sum of the first 80 odd integers. Make your own term and value chart. Label the top row, 'number of odd integers' and the bottom 'Sum'. 1 solutions
Answer 366443 by richard1234(5390) on 2012-02-02 22:04:51 (Show Source):
You can put this solution on YOUR website!Let S_n be the sum of the first n odd (positive) integers (e.g. S_n = 1+3+...+(2n-1))
n---S_n
1 1
2 4
3 9
4 16
k k^2
80 80^2 = 6400
It can be proved using induction that the sum of the first k odd integers is k^2.
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Sequences-and-series/566625: Given the sequence 1331, 1000, 729, x, y, z, 125, . . . , what is the sum of x, y, z? Please show me the steps. 1 solutions
Answer 366310 by richard1234(5390) on 2012-02-02 08:37:42 (Show Source):
You can put this solution on YOUR website!1331 = 11^3
1000 = 10^3
729 = 9^3
.
.
.
125 = 5^3
Therefore, x,y,z = 8^3, 7^3, 6^3 or 512, 343, 216. Their sum is 512+343+216 = 1071.
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Numbers_Word_Problems/566491: GIVEN THIS CONJECTURE; ANY TWO CONSECUTIVE SQUARE NUMBERS IS AN EVEN NUMBER. FIND THE COUNTER EXAMPLE 1 solutions
Answer 366224 by richard1234(5390) on 2012-02-01 21:03:27 (Show Source):
You can put this solution on YOUR website!You shouldn't type in all caps. Also, your question is ambiguous; "any two consecutive square numbers" cannot be one even number. If you stated something like "Any two consecutive square numbers *are* even numbers" then there would be infinitely many counterexamples (since the statement cannot be true), e.g. 4 and 9.
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Travel_Word_Problems/566382: what is an object's change in position relative to a reference point called 1 solutions
Answer 366223 by richard1234(5390) on 2012-02-01 20:59:11 (Show Source):
You can put this solution on YOUR website!Displacement
Note that distance is not the same thing as displacement, as distance represents the total distance travelled by a path, while displacement represents only the change in position (e.g. the shortest possible distance between two points).
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Sequences-and-series/566162: The question is: what is the sum of 4+11+18+....+4001
I have been taught to add the first and last number, so i get 4005. Then I usually divide by half of the last number, which would be 2000.5
However, I know this only works if the numbers go up by one, not seven. This is where I get lost. 1 solutions
Answer 366160 by richard1234(5390) on 2012-02-01 16:11:31 (Show Source):
You can put this solution on YOUR website!Let S = the sum you want. Then, we have
S = 4 + 11+18+...+4001
S = 4001+3994 + ... + 4
If we add them up, we get 2S = 4005 + 4005 + ... + 4005. Now we have to count the number of "4005s" there are, which is equal to the size of the set {4, 11, 18, ..., 4001}. This is equal to the size of the sets {7, 14, 21, ..., 4004} and {1, 2, 3, ..., 572} (we are just adding or multiplying numbers in the sets by a fixed number; this does not change the size of the set).
Therefore, there are 572 "4005s." Hence, 2S = 572*4005, S = 572*4005/2 = 1145430.
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Proofs/566235: Prove that the sum of any pair of rational numbers is a rational number. 1 solutions
Answer 366155 by richard1234(5390) on 2012-02-01 16:02:30 (Show Source):
You can put this solution on YOUR website!Let  and  be two rational numbers, where a,b,c,d are integers. Then,
Since integers are closed under addition and multiplication, the numerator and denominator will both be integers. Hence, the sum of two rational numbers is a rational number. We can also say that rational numbers are closed under addition.
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Functions/565931: give an example of three f,g, and h (none of which is a constant function) such that fog=foh, but g is not equal to h. 1 solutions
Answer 366021 by richard1234(5390) on 2012-01-31 23:22:00 (Show Source):
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Functions/565454: HOW DO YOU DETERMINE THE TURNING POINTS OF A GRAPH GIVEN THE FUNCTION: F(X)=2(X-3)(X^2+4)^3
1 solutions
Answer 366020 by richard1234(5390) on 2012-01-31 23:20:54 (Show Source):
You can put this solution on YOUR website!Take the derivative of f with respect to x (I used the product rule):
Set it equal to zero and find all real x that satisfy. Once you have done this, check to make sure that the sign of f'(x) actually changes.
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Polynomials-and-rational-expressions/565600: Essay: Show all work. The area of a rectangular swimming pool is given by . One side length of the pool is given by 2x + 5 feet. What is an algebraic expression for the other side length of the pool? Simplify this, and include correct units as part of your answer. 1 solutions
Answer 366018 by richard1234(5390) on 2012-01-31 23:16:13 (Show Source):
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Geometry_proofs/565708: Two circles meet at points P and Q, and diameters P A and P B
are drawn. Show that the line AB goes through the point Q. (Probably it is easier to
think of drawing the lines AQ and QB and then showing that they are actually the
same line.) 1 solutions
Answer 366017 by richard1234(5390) on 2012-01-31 23:15:06 (Show Source):
You can put this solution on YOUR website!
The easiest solution is probably to draw the segment connecting the centers of the circles (denote Y,Z), as well as segment PQ:
Since PY = (1/2)PA and PZ = (1/2)PB, triangles YPZ and APB are similar with a 1:2 ratio. Additionally, PR = (1/2)PQ (this can be proven by symmetry). Since R lies on YZ, Q must lie on AB.
Or, another way you can prove it is show that the pairs of triangles PRY/PQA and PRZ/PQB are similar. Then, you may let angle PRY = m, angle PQA = m, it follows that angle PRZ = angle PQB = 180-m. Hence, angles PQA + PQB = 180, so A,Q,B are collinear.
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