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2 years ago

Evaluate 5x-\dfrac xy5x−

Mathematics

1 answer:

olga2289 [7]2 years ago

7 0

Answer:

5x - (x/y) = 18 when x=4 and y=2

Step-by-step explanation:

5x - (x/y)

When x=4 and y=2

Substitute the value of x=4 and y=2 into the expression

5x - (x/y)

5(4) - (4/2)

Multiply 5 by 4 and divide 4 by 2

=20 - 2

Subtract 2 from 20

=18

5x - (x/y) = 18 when x=4 and y=2

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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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Answer:

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Step-by-step explanation:

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