Various Math #6 - Q&A

Question 1 Expressing the area of a square as a function of its perimeter The text is asking us to state the area A of a square as a function of the perimeter p. If x = the length of a side of a square then the area A of the square = x² The perimeter p = 4x. The solution is P² 16 I see why the solution is what it is, but I'm struggling to understand how to derive the solution algebraically. Answer 1 Perimeter p = 4x ⇒ x = P. Hence 4 2 Area a = x2 = () 2 p 42 p² 16 Answer 2 The perimeter P of a square is sum of its sides s: P = s + s + s + s = 4s The area A of a square with sides is: A = s.s=s2 Step 1: Solve s from the formula for the perimeter. Step 2: substitute s from the formula for the perimeter into the formula for the area. Step 1: P = 4s S= P 4 Step 2: A = s² S= P 4 A = P4 4 p2 A = 42 p2 A = 16 Answer 3 Let's express the area of the square in terms of its perimeter. Explanation: If s = side of the square, then, Area of the square (A) = s. s (1) Perimeter of the square (p) = 4s ⇒s= Putting this value of s in equation (1), A = ()×() P 16 Therefore, the area of a square as a function of its perimeter p is equal to . 16. Answer 4 To express the area A of a square as a function of its perimeter p, we can follow these steps algebraically: Start with the equation for the perimeter of a square: p = 4x, where x is the length of a side of the square. Solve the equation for x to express it in terms of p: x = P. 4 Substitute the value of x in the equation for the area of a square: A = x2 = ()2. 4 Simplify the expression for a by squaring (): A = p² 16 Therefore, the area A of the square can be expressed as a function of its perimeter p using the formula A = 16 Question 2 A basket of negligible weight hangs from a vertical springscale of force constant 1500 N%2Fm. If you suddenly put an adobe brick of mass 3.00 kg in the basket, find the maximum distance that the springwill stretch. Solve for the distance If the spring will stretch. Show transcribed image text A basket of negligible weight hangs from a vertical springscale of force constant 1500 N/m. If you suddenly put an adobe brick of mass3.00 kg in the basket, find the maximum distance that the springwill stretch. Solve for the distance yf the spring will stretch. Answer force in the spring is F=kx the force in the spring is opposite of the force that isexerted gravity on the brick so Force in spring= massofbrickX accelerationof gravity= 29.43Newtons 29.43 Newtons=1500 N/m*X so X(displacement of spring)= 29.43 1500 = 19.62mm Question 3 Sydney Retailing (buyer) and Troy Wholesalers (seller) enter into the following transactions. 1.1. Sydney accepts delivery of $29,000 of merchandise it purchases for resale from Troy: invoice dated May 11; terms 3/10, n/90; FOB shipping point. The goods cost Troy $19,430. Sydney pays $655 cash to Express Shipping for delivery charges on the merchandise. 1.2. Sydney returns $1,300 of the $29,000 of goods to Troy, who receives them the same day and restores them to its inventory. The returned goods had cost Troy $871. 2.0 Sydney pays Troy for the amount owed. Troy receives the cash immediately. (Both Sydney and Troy use a perpetual inventory system and the gross method.) Prepare journal entries that Sydney Retailing (buyer) records for these three transactions. Prepare journal entries that Troy Wholesalers (seller) records for these three transactions. Sydney pays Troy for the amount owed. Troy receives the cash immediately. Note: Enter debits before credits. Answer Step 1 Accounts Payable balance: $29000 - $1300 = $29,700 Discount received on early payments: $29, 700-3% = $831 Journal entries that sydney (buyer) records for these transactions Date General Journal Debit $ Credit $ May - 11 Merchandise Inventory $29,000 May-11 Accounts Payable $29,000 May - 11 Merchandise inventory $655 May-11 Cash $655 May - 12 Accounts Payable $1,300 May-12 Merchandise inventory $1,300 May - 20 Accounts Payable $ 27,700 May-20 Merchandise Inventory $ 831 May-20 Cash $ 26,869 Journal entries for troy (seller) to record these transactions Date General Journal Debit $ Credit $ May - 11 Accounts Receivables $ 29,000 May-11 Sales $ 29,000 May - 11 Cost of goods sold $ 19,430 May-11 Merchandise inventory $ 19,430 May-12 Sales returns and allowances $1,300 May - 12 Accounts receivable $1,300 May-12 Merchandise Inventory $ 871 May-12 Cost of goods sold $ 871 May - 20 Cash $ 26,869 May-20 Merchandise Inventory $ 831 May-20 Accounts Payable $ 27,700 Question 4 Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at T∞ = 1200°C. and maintains a convection coefficient of h = 250 W K over the blade. m² The blades, which are fabricated from Inconel, k≈ 20 W K, have a length of L=50 mm. The m blade profile has a uniform cross-sectional area of Ac = 6 x 10-4m² and a perimeter of P=110 mm. A proposed blade-cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of T₁ = 300°С. a) If the maximum allowable blade temperature is 1050°C and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant? Answer 1 Given: W W L = 0.05m, Ac = 6x10-4, P = 0.11m, k = 20-K, h = 250-2K,Tb b = 573K,T∞ = 1473K, T (L) = 13 m m² maz Schematic of turbine blade mounted to a rorating disc as follows: Show description