course Mth 271 0.3.22 (was 0.3.24 simplify z^-3 (3z^4)Պ|˱~ҼwîN}x
......!!!!!!!!...................................
11:50:55 Section 0.1.26 solve x/2-x/3>5
......!!!!!!!!...................................
RESPONSE --> mulitply by the least common denomenator 3x - 2x = 5\6 x = 5\6 confidence assessment: 1 I hate fractions
.................................................
......!!!!!!!!...................................
11:51:54 It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6: 6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you x > 6 * 5 which simplifies to x > 30. The interval associated with this solution is 30 < x < infinity, or (30, infinity). To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**
......!!!!!!!!...................................
RESPONSE --> I knew I should have multiplied, but I'm never shure what to do, I knew this one. self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:59:40 Section 0.1.28 solve 2x^2+1<9x-3
......!!!!!!!!...................................
RESPONSE --> I started by subtracting one on both sides, then square rooting both sides 2x< 3x - 2 then add two to both sides, then subtract 2x from both sides 2 < x confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:01:45 The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0. The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ] The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign. Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true. Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals. In any case the correct interval is 1/2 < x < 4. ALTERNATE BUT EQUIVALENT EXPLANATION: The way to solve this is to rearrange the equation to get 2 x^2 - 9 x + 4< 0. The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails. The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign. If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2). The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4). It changes sign again at 4 so is positive on the interval (4, infinity). The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **
......!!!!!!!!...................................
RESPONSE --> I don't understand that one... I hate inequalities self critique assessment: 3
.................................................
}_߽zoΕ assignment #001 001. `Query 1 Applied Calculus I 07-19-2007 |sO[jׅtzиz assignment #003 003. `query 3 Applied Calculus I 07-19-2007
......!!!!!!!!...................................
12:30:31 0.3.22 (was 0.3.24 simplify z^-3 (3z^4)
......!!!!!!!!...................................
RESPONSE --> multiply the exponets, which is really adding -3 to 4, equaling 3z confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:30:40 z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:36:10 0.3.28 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
......!!!!!!!!...................................
RESPONSE --> simplify the constants, (4s^2 / 3s)^3, then ditibute the exponent, 64 s^5/ 27 s^3, then divide, 2.37 s^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:37:58 Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. **
......!!!!!!!!...................................
RESPONSE --> ah, ha, I didnt simplify enough in my first step self critique assessment:
.................................................
......!!!!!!!!...................................
12:42:02 0.3.34 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
......!!!!!!!!...................................
RESPONSE --> 34. a is ((3x^2 y^3)^4)^(1/4) which would cancel the root and the square leaving 3x^2 y^3 b starts by cancelling the exponets and roots leaving 54x ^4
.................................................
......!!!!!!!!...................................
12:43:22 To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) Alternative explanation: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). **
......!!!!!!!!...................................
RESPONSE --> i dont get your explanation self critique assessment:
.................................................
......!!!!!!!!...................................
12:45:23 0.3.62 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
......!!!!!!!!...................................
RESPONSE --> all thats left is x^(n+1) confidence assessment:
.................................................
......!!!!!!!!...................................
12:45:41 Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
}ċ싔և assignment #003 003. `query 3 Applied Calculus I 07-19-2007 "