Algebra1 HELP! For today?
1. When opened the money box which yielded $ 3.67 in Nickles & pence. If there were 147 Nickles & ASSEMBLY cents, how many of each were in the bank? Consecutive odd 2.Find intergers four 5times such that the sum the first and third 25 ahead of us 5times room. 3.The velvet bag containing pieces 12gold and 8silver coin. A room is chosen at random from the bag ¬ replaced. The money is randomly extracted Secound. What is the Bothy probablibility coins are of gold? 4. If f (x) = x (squared) 3 x 1 to find f (a-1) 5.simplify: (0.00042 X 10 (4 square)) (3000 (five squared x 10)) / (70000) (0.0006 x 10 (squared )) 6. (3 + x) (x (squared) + 3x + 4) HELP !!!!!!!!!!!!! For now! PLEASE! I heart you'll always !!!!!!!!!!!!!!!!!!
1. Let n = number of Nickles let p = number of rooms we have 2 equations here: p + n = 147, 05 n + .01 p = 3.67 solve an equation in terms a variable, and replacing this value in other equations. So: to isolate n + n p = 147 p subtracted from both sides, giving n = 147-p Plunk now in the second equation, .05 (147-p) + 01 p = 3.67 multiplied are obtained 05P + .01 -. 7.35 p = 3.68 combine like terms, we obtain -. 04p = -3.68 for p = 92 However, since the original formula of n = 147 – PN n = 147-92 = 55 Thus, it is 55 nickels, 92 cents 2. Again, since variables. The same number a, b, c, d we know that 5 (a + c) = 25 + 5d you also know that a + c = 4 and d = 6 + Now, take 5 (a + c) = 25 5 D, we can be simplified if we divide both sides by 5, leaving a + c + d = 5 Substitute in values of C and D and ends with A + A + 2 = 5 + a 6 From now, I'm sure you can solve it:) I hope that 2, help you on your way!
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