Can someone Solve: 8+4m=9m-7?
What you're using is called The Property of Equality. It says that anything you do to both sides of the equal sign leaves the equation correct. In other words, if a = b, then a + c = b + c a - c = b - c ac = bc a/c = b/c As long as you do the same thing to both sides, its alright. In your equation 8 + 4m = 9m - 7 add 7 to both sides. That will cancel the -7 on the right-hand side, but make the 8 on the left-hand side a 15 15 + 4m = 9m Subtract 4m from both sides. That will cancel the 4m on the left but make the 9m on the right into a 5m. 15 = 5m Divide both sides by 5. That will make the 5m into one m, and the 15 into a 3. 3 = m If you want to check the answer, substitute 3 for m in the original equation. 8 + 4(3) = 9(3) - 7 8 + 12 = 27 - 7 20 = 20 It works. ANSWER m = 3
Can someone solve (2a+4)(4a+7)?
Hi, For this question, we have to use the FOIL Method to multiply these two quantities. Recall, that when using this method on this particular example, we should get the following: [2a][4a] + [2a][7] + [4][4a] + [4][7] Now, complete the above multiplication to get the following: 8a^2 + 14a + 16a + 28 Simplifying the above gives us the following: 8a^2 + 30a + 28 <==== FINAL ANSWER I hope that helps you out! Please let me know if you have any other questions!
Can someone solve this set?
There's already a rigorous answer, I'll try to view it intuitively (though it can be somewhat translated to a more set theoretic approach than just text arguments).The set [math]A-B[/math] is the set of elements in A that is not in B, while the set [math]A+B[/math] is the set of elements that is either in A or B but not both.Statement 1: True. If [math]A+B=B[/math], that means A is actually empty: if A is non-empty, consider a member of A. If it's not in B, then it'll be in [math]A+B[/math]. But that implies it is in B, contradiction. If it's in [math]B[/math], that means it is in [math]A+B[/math]. But since it's in both A and B, it doesn't belong in [math]A+B[/math], contradiction. Hence, A is empty -- but the empty set is a subset of any sets, so [math]A \subseteq B[/math].Statement 2: True. If there's no element that belongs exclusively in A or in B, that means any element in A is also in B, and vice versa. Hence, they are both equal.Statement 3: True. If the set of elements that is exclusively in A or B is equal to the union of A and B, that means there's no element that is in both -- i.e. the intersection of A and B is empty.
Can someone solve x^x=e?
The solution for [math]x^x = e[/math] is a root of [math]f(x) = x ln (x) - 1[/math], which is a smooth curve with a single root and and thus a good candidate for the Newton-Raphson method. For simple functions such as this, it’s quite easy to compute using any computer language or even a calculator.As the function is smooth and well-behaved, the method converges and finds the root (solution) very fast. I’ll show the steps below.The equation for the Newton-Raphson method is [math]x_{n+1} = x_n - f(x_n) / f'(x_n)[/math]If you plug in [math]f(x) = x ln (x) - 1[/math] and its derivative [math]f'(x) = ln (x) + 1[/math] and start with [math]x_0 = 1[/math], you get:[math]x_0 = 1 [/math][math]x_1 = 2.0[/math][math]x_2 = 1.7718483[/math][math]x_3 = 1.7632362[/math][math]x_4 = 1.7632228[/math][math]x_5 = 1.7632228[/math]I calculated the above on a Lisp REPL, but you could use, say, Python, Javascript or even a pocket calculator and pen and paper. Either way the quadratic convergence is pretty good and yields 6 decimal places after only 4 iterations.
Can Someone Solve This?
Molar mass of NaN₃ = (23.0 + 14.0×3) g/mol = 65.0 g/mol The equation for the reaction : 2NaN₃ → 2Na + 3N₂ Mole ratio NaN₃ : N₂ = 2 : 3 No. of moles of NaN₃ = (125 g) / (65.0 g/mol) = 1.923 mol No. of moles of N₂ gas produced = (1.923 mol) × (3/2) = 2.885 mol For the N₂ gas produced : Pressure, P = 756/760 atm No. of moles, n = 2.885 mol Gas constant, R = 0.08206 L atm / (mol K) Temperature, T = (273 + 27) K = 300 K Gas law : PV = nRT Volume, V = nRT/P = 2.885 × 0.08206 × 300 / (756/760) L = 71.4 L
Can someone solve this?
moles of sodium oxalate = 1.0423 / 134.00 = 0.00778 these were dissolved in 250 ml and a 25 ml aliquot taken so moles of oxalate in the 25 ml = 0.00778/10 or 7.78 *10^-4 now we need an equation 2 MnO4- + 5 C2O4 2- + 16 H+ >> 2 Mn2+ + 10 CO2 + 8 H2O 1 mole of the oxalate reacts with 2/5 moles of the permanganate moles of the MnO4- = (7.78 *10^-4) *2/5 = 3.1*10^-4 these moles are in 28.93 mL so the moles per liter = 0.00031 * 1000/28.93 0.0107 moles per liter MnO4-
Can someone solve this?
You can sell 100 pet chias per week if they are marked at $1 each, but only 30 each week if they are marked at $2/chia. Your chia supplier is prepared to sell you 10 chias each week if they are marked at $1/chia, and 90 each week if they are marked at $2 per chia. demand function q (p) = supply function q (p) = At what price should the chias be marked so that there is neither a surplus nor a shortage of chias?:
Can Someone solve this?
For the first pilot, there are 3 fish dishes out of 6 available. For the second pilot, there are 2 fish dishes out of 5 available. 3/6 x 2/5 = 1/2 x 2/5 = 1/5 Answer: 1/5
Can someone solve this problem?
First, put it in 'standard' quadratic form 2x² - 16 x - 24 = 0 Since the x² term is positive, the parabola opens 'upwards' and the vertex will be 'centered' between the roots (where it crosses the x-axia). So, solve it (using the quadratic formula since it won't factor 'nicely') and then calculate half the distance along the x-axis from crossing to crossing. Add that value to the 'left-hand' crossing to get the x-coordinate of the vertex. Then plug the x coordinate into 2x² - 16 x - 25 and do the arithmetic to get the y-coordinate of the vertex. The domain of the function is simply, in this case, all of the reals from -∞ to +∞, and the range is from the y-coordinate of the vertex to +∞. *You* do the arithmetic so you'll understand it better. HTH ☺ Doug Edit: Oops... Looks as if I forgot to RTFQ (Read The F*ckin' Question ☺) Oh well... It's not the first time I've made an αsshole out of myself, and it probaby won't be the last ☺
Can someone solve these?
I think this will satisfy u