Distance travelled on level road = (S + 20 ) time = (S + 20 )x 6 Where S = Speed on winding road Distance travelled on the winding road = Speed x time =S x3 Total distance travelled on two rods = 6(S +2) +3S =516 9S +12 = 516 or 9S =504 S...

Distance travelled on level road = (S + 20 ) time = (S + 20 )x 6 Where S = Speed on winding road Distance travelled on the winding road = Speed x time =S x3 Total distance travelled on two rods = 6(S +2) +3S =516 9S +12 = 516 or 9S =504 S...

Mass of water =3.35 Kg Temperature of water = 20C Mass of Ice =0.4 Kg Temp of ice = -10.2C Heat lost by water till it reaches 0C =MSt = 3.35*4.2*103*20 J Heat gained by ice till it reaches 0C = MSt=0.4*2.1*103*10.2 J Difference=A?=Difference of above two values Mass...

What are the order pairs to -3x+8=y (answer)

-3X+8=Y let X be zero Y =-8 So the ordered pair is 0,-8 likewise if Y= 0 , X =8/3 so pair is 8/3, 0

Volume of a box. (answer)

Height = VolumexLength xBreadth Now plug in the given values

Find the discount rate when the original price is $101 and the amount of the discount is $20.20 (answer)

Percent discount = Discount x 100/ Origional price

solve and put in a trig equation (answer)

25 Cos 2t +12 Sin (t+pi/2) + 2 Cos2 t = 25 Cos2 t - 25 Sin2 t + 12 Cos t + 2 Cos2 t = 27 Cos 2 t -25 Sin2 t + 12 Cos t = 52 Cos2 t -25 + 12 Cos t

The lengths of sections AB , BC , CD and DE are 12x, 5x , 4x and 5x and Total length of four together = 52 cm so 26x = 52 x= 2 Thus AB = 24 cm BC = 10cm CD = 8Cm and DE = 10cm The height of trapezium is 6cm So area = base...

Math question involving profits (answer)

Let X = Normal selling price Price after discount = X - 0.2X =0.8 X Ley Y be the buying price of trader The profit originally was X-Y New Profit = 0.8X- Y =0.04 (X-Y) 0.76X = 0.96Y X/Y = 1.263 (X-Y)/ Y is fractional profit multiplied by 100 will give original percent...

Let x = the number 5(X+6) = X-2 5X+30=X-2 4X =-32 X= -8

Chemistry Help, Boyle's Law? (answer)

P1V1= P2V2 P1 = 1 at V1 = 2* 107 P2= 23 at V2 = P1 V1/ P2 Compression = Difference in volumes

There are three forces acting on each mass F1 = G m1*m2/ d2 and F2 = Gm2*m3/d2 and F3 = G m4*m2/ 2d2 As diagonal distance is √2 d Here F1 and F2 are perpendicular to each other and F3 is 45 degrees to both F1 and F2 Using vector...

Let H = total number of horses C = Total number of cows H + C = 995 h = number of horses sold and c = Number of cows sold Number of horses left = H- h Number of Cows left = C-c H- h = 510.... Equation 1, Gives h =H -510... Equation A and C...

Volume of liquid in A = 2/2 = 1 litre Volume of liquid in B = (1/3)*3 = 1 litre Volume 0f liquid in C = zero Total volume = 2 Litre This is divided equally in three parts Each part = 2/3 Lt Capacity of C is 6 Lt but it has only 2/3 Lt Fraction of C filled...

Rewrite as the given equation as Y = - 3X +4 Using Y = mX +b where m = slope and b = Y intercept m = -3 and b = 4

tan(theta) =8/6 = 4/3 Where theta is the angle with horizontal when x = 6m = y/x(as y= 8m when x= 6 m) y=4/3(x) dy/dt =( 4/3) dx/dt Speed on wall =( 4/3 ) speed on ground = (4/3) 0.5 =2/3 m/s = Speed on the wall

The perimeter of the quadralateral is 57 cm. Find the lengh of each side x+5, x+7, x^2-3x,3x-3 (answer)

Perimeter = (X+5) +(X+7) +(X2-3x) +(3X -3) = 57 Gives X2 +2X+9= 57 X2 +2X -48=0 X2 +8X -6X -48 =0 X(X+8) -6 (X+8) Gives X = 6 and X = -8 Now use X = 6 to find X+5 , X+7, X2-3X and 3X-3 as the sides of q...

II is the case of joint variation as Area A depends on l and w In I and II it depends on s and r respectively as pi is constant

Original Price = 3.99 +6.99 + 5.99 = 16.97 Sale price = 14.50 ( difference) Saved = 16.97 - 14.50 = 2.47 Percentage saving = 2.47 *100/ 16.97 = 14.56 % For Second part Let X = Original price Saving = X*30/100 Payable...

Let X = Total number of Votes 5/8 X = Blue 5/9 X = Green Remaining = X - ( 5/8X +5/9X ) =48 = Red Some thing wrong with the data as X can not be negative

Hello 45X + 2000 = Y Where X = Number of units Y = Total Cost