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Kazeer [188]
3 years ago
5

Devorah is filling a pool with a hose. The volume.H. In liters, of water coming out of the hose in .m.minutes is given by the fu

nction H(m)=17.4m. However it is a sunny day, and water is also evaporating from the pool. Therefore,the volume ,V, in liters, of water in the pool m minutes after devorah started filling it is given by V(m)=17m.
Let E be the volume of water, In Liters ,that has evaporated from the pool m minutes after devorah started filling it .

Write a formula for E(m) in terms of H(m) and V(m)

E(m)=___________

And

Write formula for E(m) in terms of m.

E(m)=______________

Mathematics
2 answers:
Alja [10]3 years ago
8 0
<span>Given that Devorah is filling a pool with a hose. The volume.H. In liters, of water coming out of the hose in .m.minutes is given by the function H(m)=17.4m. However it is a sunny day, and water is also evaporating from the pool. Therefore,the volume ,V, in liters, of water in the pool m minutes after devorah started filling it is given by V(m)=17m.

  IfE be the volume of water, In Liters ,that has evaporated from the pool m minutes after devorah started filling it .

The formula for E(m) in terms of H(m) and V(m) is given by

E(m) = H(m) - V(m)

And

The formula for E(m) in terms of m is given by

E(m) = 17.4m - 17m = 0.4m</span>
bulgar [2K]3 years ago
8 0

Answer:

H(m)-V(m)

0.4m

Step-by-step explanation:

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gladu [14]

Option C:

\frac{3 a^{2} b^{11}}{2 } is equivalent to the given expression.

Solution:

Given expression:

$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}

To find which expression is equivalent to the given expression.

$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}

Using exponent rule: \frac{1}{a^m}=a^{-m}, \ \  \frac{1}{a^{-m}}=a^{m}

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    $=\frac{-18 a^{-2} a^{4} b^{5} b^{6}}{-12 }

Using exponent rule: {a^m}\cdot{a^n}=a^{m+n}

    $=\frac{-18 a^{(-2+4)} b^{(5+6)}}{-12 }

   $=\frac{-18 a^{2} b^{11}}{-12 }

Divide both numerator and denominator by the common factor –6.

   $=\frac{3 a^{2} b^{11}}{2 }

$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}=\frac{3 a^{2} b^{11}}{2 }

Therefore, \frac{3 a^{2} b^{11}}{2 } is equivalent to the given expression.

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