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Is the cube root of 5 irrational?
Yes it is. There is no fraction which, when cubed, equates to 5. Consider n = p/q where p and q are integers expressed in lowest terms (they are relatively prime). If n3 = 5, then p3 / q3 = 5. This equates to an integer if and only if q3 = 1 meaning q = 1 or if p3 is divisible by q3. The latter is impossible since p and q are relatively prime. Thus, for n to be the cube root of 5 and be rational, it must be an integer. No integer cubes to 5 (1, 8, ...). Thus, the cube root of 5 is irrational.
How do you find a growth rate of 1.14?
Q(t) = Q0(1.14t) Q = the quantity at time t t = time (in, say, years) Q0 = the quantity at time zero For example, if t = 5 years and Q0 =1000, Q(5) = 1000(1.145) = 1000x(1.925414...) = 1925.414... = about 1925.
How do you write q2 and verbal expression?
Write an algebraic expression for the verbal expression. q squared minus 2 times q
What is the quotient and reminder for 113 divided by 5?
22.6
James has 33 coins in his pocket all of them nickels and quarters If he has a total of 2.65 how many quarters does he have?
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
