Need help in scientific notation and significant figures?
Given the equation w = xyz, and x = 1.7 · 103, y = 2.23 · 10-2, and z = 8.000, what is w, in scientific notation and with the correct number of significant figures? I tried multiplying the mantissas and adding the exponents but for some reason I can't come up with the right answer. Well, that's what my online thing says... So if anyone can help, it would be greatly appreciated. thanks for your help
I have several questions regarding significant figures and exponents as well, I appreciate any help you can send my way! thanks!?
(1) Calculate with all the precision available, then round to the appropriate number of significant figures. (The "exact" numerator product is actually 11.340*10^-3. The "exact" denominator product has 5 significant digits as well.) Your answer will be 1.8*10^1, or 18. (2) 10^1 is not 1. It is 10. See above for answer choices. You might pick 1.8*10^1 if you are requested to put the answer in scientific notation. It makes more sense to me to write it as 18, as this uses less ink and is directly understood as a number with 2 significant digits. (If the answer has trailing zeros, scientific notation works well to communicate the number of significant digits.) For your own purposes (as opposed to the purposes of your math teacher), you may find it occasionally convenient to use a scientific notation form with a zero to the left of the decimal point. I like to use this for numbers in a denominator, as their inverse is then a number in the range 1-10 (before taking exponents into account). This can make it easier to estimate answers. Thus, if my next calculation used the result of this one in the denominator, I might write it as (numerator)/(0.18*10^2), which I could estimate as 5.6*numerator*10^-2. (Note the use of parentheses to show the extent of the numerator and denominator.)
Do i round off three significant digits and express the result in standard scientific notation 254,931?
There is no rule that states that numbers are always expressed to 3 significant figures in scientific notation. The degree of rounding depends on the other data in the question . If you have to express your final answer to 3 significant figures , because the least precise measurement that was used in the calculation had 3 significant figures, then the final answer will be: 2.55*10^5 But you can equally express this to 2 sig figs: 2.5*10^5 Or to 4 sig figs: 2.549*10^5
I'm confused with significant digits. How many in 5.000? 4.9 x 10^3? Rules, explanations & examples, please.?
All significant figures count from the first non zero number regardless of any other rules. If there is no decimal point then all the zeros after the last non-zero number are NOT significant. If there is a decimal point then all the zeros after the first non-zero number ARE significant. 5.000 = 4 sig figures. (all zeros after first non-zero count because of decimal point) 5000 = 1 sig fig, no decimal point 0.0005 = 1 sig figure, only number after first non-zero count 0.00050 = 2 sig figures 5.0005 = 5 sig figures If there is no decimal point then a zero after the last non-zero number is not counted as significant 500 = 1 sig figure 500.0 = 4 sig figures (has a decimal point, which makes the zero after the point significant and thus all the others as well) 501 = 3 sig figures For 4.9 x 10^3 it is 2 sig figures, apply the same rules as above to the numbers in the scientific notation. Writing it out long hand you get 4900, and as above this is two sig figs 0.3 x 10-3 one sig figure long hand 0.0003 1 sig figure Hope this helps, I have put a couple of links below
Can someone help me with scientific notation problems from chemistry?
there are some basic rules for example Rule 1 10^ a x 10^ b = 10^(a + b) eg..... 10^ 3 x 10^ 4 = 10^( 3 + 4 ) = 10^ 7 Rule 2 10^ a / 10^ b = 10^(a - b) eg..... 10^ 5 x 10^ 3 = 10^( 5 - 3 ) = 10^ 2 or 1 / 10^ 4 = 10^ -4 Rule 3 if u have a very big value u can right it in the Scientific notation form ... having the base 10 eg .. 25000000. its big value.... Scientific notation makes big value small and vice versa 2.5 x 10^ 7 u can see in above value decimal point is at end u moved the decimal point before the first digit ... now count the digit from where u move the decimal point in this case u move decimal point to 7 digit so u placed the power 7 on the base 10 same way it can b done for very small value 0.00000012 = 1.2 x 10^ - 7 1/0.00032 = 3.125 10^3/10^-3 = 10^ [ 3 - (-3) ] = 10^ [ 3 + 3 ] = 10^ 6 Rule 2 1/55,000 = 1 / ( 5.5 x 10^ 4) rule 3 = 0.1818 x 10^ -4 = 1.818 x 10^ -3 (10^5)(10^4)(10^-4) / (10^-2) = 10^ [ 5 + 4 -4 - (-2) ] = 10^ ( 5 + 2 ) = 10^ 7
What is an intuitive explanation for the process of exponentiation?
Imagine you have a function with the following property: [math]f(x+y) = f(x)f(y)[/math] for all [math]x,y[/math] in reals. What can you say about that function?It's easy to verify the following properties:[math]f(0) = 1[/math] because [math]f(x) = f(x+0) = f(x)f(0) \implies f(0) = 1[/math][math]f(nx) = f(x+x+\dots+x) = f(x)f(x)\dots f(x) = f(x)^n[/math]It's fairly easy to see that, when you restrict [math]x[/math] to integers, then [math]f(x) = a^x[/math] for any [math]a>0[/math].Your question then becomes, what does it mean for [math]f(a/b)[/math]?Look at our defining property: [math]f(x+y) = f(x)f(y)[/math], and set [math]x,y = 1/2[/math], to get [math] a = f(1) = f(1/2 + 1/2) = f(1/2)f(1/2) = f(1/2)^2 = f(2*1/2) = f(1) = a[/math]. So [math]f(1/2)[/math] is the number that, when raised to the 2nd power, is [math]f(1)[/math]. In other words, the square root of [math]f(1)[/math].The same argument can be made for any reciprocal argument: [math]f(1) = f(n/n) = f(n\times1/n) = f(1/n)^n \implies f(1/n) = \sqrt[n]{f(1)}[/math].This can be extended to any rational number: [math]f(a/b) = f(1/b)^a = \sqrt[b]{f(1)}^a[/math]. Does this make sense to you?To round things out, let's also look at negative arguments: [math]1 = f(0) = f(a-a) = f(a)f(-a) \implies f(-a) = \frac{1}{f(a)}[/math]Or, equivalently, [math]f(-a) = f(-1 \times a) = f(a)^{-1}[/math]. It's nice that the two ways of computing [math]f(-a)[/math] give the same value. It gives us confidence that what we are doing makes sense.So the whole reason why rational powers of exponents are defined the way they are is because of a strong desire to preserve the very important relationship [math]a^{x+y} = a^xa^y[/math].Extending this to real exponents is slightly more tricky, and to do it properly requires picking a definition of reals. The basic idea is that if [math]x \approx a/b[/math], then [math]c^x \approx c^{a/b}[/math], and so if [math]\lim_{n\to\infty} a_n/b_n = x, c^x = \lim_{n\to\infty} c^{a_n/b_n}[/math]. For extending it to complex numbers, it helps to use the identity [math]e^x = x^0/0! + x^1/1! + x^2/2! \dots[/math], which can be shown to have the desired properties of exponentiation. In which case, [math]e^i = i^0/0! + i^1/1! + i^2+2! + \dots = 1+i-1/2-i/6+\dots \approx 0.54+0.84i[/math].
Use induction to prove 5^2n - 1 is a multiple of 8 for all natural numbers n.?
I'll first switch notation slightly to make things easier, and I'll do this by using the laws of exponents. Note that x^(ab) = (x^a)^b. So that means 5^2n = (5^2)^n = 25^n. So 5^2n - 1 = 25^n - 1, and we are to show this is divisible by 8 for all natural numbers n. Well, first we must show it to be true for a base case. Since we're to show it's true for all natural numbers, that means we must show it's true for n = 1. But 25^1 - 1 = 25 - 1 = 24, and we know 24 is a multiple of 8. So true for n = 1. Next we assume it to be true for n = k, so that 25^k - 1 is divisible by 8. That means there is some integer, which we shall call a, such that 25^k - 1 = 8a. And therefore 25^k = 8a + 1. Next we need to use this assumption to show it to be true for n = k+1, in other words 25^(k+1) - 1 is divisible by 8. That means we need to write it in the form of 8 times some integer. But 25^(k+1) - 1 = (25^k)(25^1) - 1 by the property of addition of exponents = 25(25^k) - 1 = 25(8a+1) - 1 using the assumption = 200a + 25 - 1 expanding out the brackets = 200a + 24 = 8(25a + 3) which is in the form of 8 times an integer and is therefore divisible by 8.
Lords above, help me! (chemistry questions)?
Okay, so I am making a well meaning attempt to not fail my chemistry homework. My school thought it would be an awesome idea to give us homework through the University of Texas, so of course it's ridiculously hard. My grade currenlty hangs in the balance, any help is appreciated! Question 1: A bathroom scale can weigh something up to 200 lbs. The 200 lbs is divided into one pound increments. A boy's weight falls between 113 and 114 pounds on this scale. How many significant figures can be reported in this weight? 1. 5 2. 3 3. 4 4. 2 Question 2: What is the proper solution to the following equation? (150 + 2.5 + 36.75) divided by (6.60 + .173) 1. 27.94 2. 28 3. 27.9418 4. 27.94183 5. 30 6. 27.942 7. 27.9 Question 3: Perform the operation - (1.65 cg) + (.231 g) *this is like factor labeling, you have to change cg to grams. Question 4: Perform this operation - (7.322 x 10e9 cg) - (7.07 x 10e7 g) *same thing, and the "e" means exponent. Sorry, my computer doesn't have the little numbers for it :) Question 5: (.00083 m) x (2.75 cm) answer in units of meters squared Question 6: Perform the operation - .00319 cg divided by 68900 mL answer in units of g/mL Question 6: Which of the following numbers has the greatest number of significant figures? 1. 0.0032 2. 4.3 x 10e2 3. 1.2 4. 130 5. 0.690 6. 33,000 Question 7: How many significant figures would be retained in the solution to- [(2 + 0.0031) + (413/12)] x (2.45 x 1.83) 1. five 2. one 3. two 4. zero 5. three 6. four Any help is sooo appreciated. I'm usually not this dumb, but chemistry holds no appeal to me and I'm awful at it. The main theme of this homework is significant figures, in case you didn't catch that. Thanks so much, to everyone smarter than me out there!