Math Help

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1

<p>Can someone help me with some of these problems and explain how to do them. Thanks in advance…</p>

<li> (x-a)^5 = (x-a)k^4<br></li>
</ol>

<p>In the equation above, x and k are positive numbers and 0 < a < x. Which of the following must be equal to X???</p>

<p>a) k
b) k-a
c) k+a
d) a^4
e) k^4 + a</p>

<li>The five digits 1,2,3,4 and 5 are used to form five-digit numbers greater than 40,000 are possible?? (I can do these if I waste like 5 minutes on them…but I know there is an easier way to do this).</li>
</ol>

<p>a) 24
b) 48
c) 64
d) 96
e) 120</p>

<li>If (a-10b)^2 + 100b^2, what is the value of a^2*b^4??</li>
</ol>

<p>More might come…</p>

<p>THanks guys…</p>

2

<p>hey i know the first one...not too sure about the 2nd..</p>

<p>ok..since x and k are positive numbers, i set k equal to 1 and x equal to 2. For a, i set it equal to 1 since it had to be less than 2 (which is x).. then i substituted all of the values into the equation just to check that both sides are equal. To find the value of x, i just substituted each multiple choice answer with the values that i set. C was my final answer. On problems like these, you can substitute any numbers as long as the follow the question and substitute the multipe choice answers in.</p>

3

<p>For the second one, I'm assuming that the question is how many numbers are possible, and i'm assuming that "distinct" is thrown into the question.</p>

<p>2x4x3x2x1=48</p>

4

<p>Ya, number one is C. I just substituted a value for a and x, and just solved for k. Then I used the value of k and a to find x.</p>

<p>(x-a)^5=(x-a)k^4
(5-1)^5=(5-1)k^4
4^5=4k^4
1024=4k^4
256=k^4
k=4</p>

<p>Remember x=5</p>

<p>a. 4=4 not 5
b. 4-1=3 not 5
c. 4+1=5 yes 5</p>

5

<p>surge can you explain the formula you applied to get those numbers..</p>

6

<p>I think it's much easier to do number 1 algebraically, plus it saves more time: </p>

<ol>
<li>(x-a)^5 = (x-a)k^4</li>
</ol>

<p>divide both sides by (x-a), and get</p>

<p>(x-a)^4 = k^4
taking the fourth root of both sides yields
x-a = k
x = k+a</p>

7

<p>"surge can you explain the formula you applied to get those numbers.."</p>

<p>Yeah I was wondering the same thing....Well now I know how to do the first one (long way and short way---After seeing how to do it Algebraically, I feel very stupid) Thanks guys. </p>

<p>Now Number 2 and 3.</p>

8

<p>my method takes < 30 seconds, I just put all the steps for him to follow.</p>

9

<p>There is a really fast method for number 2.</p>

<ol>
<li>The five digits 1,2,3,4 and 5 are used to form five-digit numbers greater than 40,000 are possible?? (I can do these if I waste like 5 minutes on them...but I know there is an easier way to do this).</li>
</ol>

<p>Now, we know that 3 of these numbers, when made the initial number, (1,2,3) will make the number less than 40,000.</p>

<p>This means 3/5 of the possible numbers will make combinations <40,000.</p>

<p>Use a graphing calculator to find 5! (Five Factorial, which is the number of all possible combinations).</p>

<p>3/5 of the combinations (1,2,3) will be <40,000 when made the first digit, which puts them out of the running for possible combinations</p>

<p>2/5 will be greater than 40,000.</p>

<p>5!=120</p>

<p>120 x 2/5 = 48</p>

<p>QED</p>

10

<p>Here is my method for #1:</p>

<p>(x-a)^5 = (x-a)k^4</p>

<p>This obviously means that k^4 = (x-a)^4, because it would be multiplied once more by first (x-a) on the right side to = (x-a)^5, which would make the equation true.</p>

<p>Now, if k^4=(x-a)^4, then k=x-a (take the 4th root of each)</p>

<p>Now, simply isolate the x. Inverse operations tells you that x=k+a</p>

<p>*all other methods are also very good</p>

<p>QED</p>