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Len [333]
2 years ago
9

Ralph is three times as old is Sarah. In six years Ralph will be only twice as old as Sarah will be then. Find Ralphs age now. W

hich of the following equations can be used to solve the problem.
Mathematics
2 answers:
andrew11 [14]2 years ago
5 0
Let's begin by calling Sarah's age now as X. As Ralph is 3 times as old as Sarah, X times 3 = 3X. Hence, Ralph's age is 3X. In six years, Ralph will be twice as old as Sarah. To calculate six years from now, add 6 to X for Sarah, and 6 to 3X for Ralph. As Ralph is twice as old as Sarah and we want to find the difference between the ages to calculate X, multiply X+6 by 2. You'll get 2X+12. Therefore, 2X+12=3X+6. Deduct 6 from 3X+6 as we want to isolate the variable. Because you did that to one side, you have to deduct 6 from 2X+12. Hence, now you have 2X+6=3X. X=6. Ralph's age is 3X, so 6 times 3 is 18. Ralph is 18 years old. 
luda_lava [24]2 years ago
3 0

Answer:3x+6=2(x+6)

Step-by-step explanation:

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A trough has ends shaped like isosceles triangles, with width 5 m and height 7 m, and the trough is 12 m long. Water is being pu
Svet_ta [14]

Answer:

\dfrac{dh}{dt}=21 \text{m/min}

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

V = \dfrac{1}{2}(bh)\times L

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

\dfrac{b}{h} = \dfrac{5}{7}

b=\dfrac{5}{7}h this can be substituted back in the volume equation

V = \dfrac{5}{14}h^2L

the rate of the water flowing in is:

\dfrac{dV}{dt} = 6

The question is asking for the rate of change of height (m/min) hence that can be denoted as: \frac{dh}{dt}

Using the chainrule:

\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

V = \dfrac{5}{14}h^2L

\dfrac{dV}{dh} = \dfrac{5}{7}hL

reciprocating

\dfrac{dh}{dV} = \dfrac{7}{5hL}

plugging everything in the chain rule equation:

\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}

\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6

\dfrac{dh}{dt}=\dfrac{42}{5hL}

L = 12, and h = 1 (when the water is 1m deep)

\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}

\dfrac{dh}{dt}=21 \text{m/min}

the rate of change of height when the water is 1 meter deep is 21 m/min

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What you do first is add the books 42 + 85 + 69 which equals 196. Then divide 196 by 7 and you get 28 :)
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