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olya-2409 [2.1K]
2 years ago
10

A³b² a²b simplify the following expression

Mathematics
1 answer:
liberstina [14]2 years ago
7 0

Answer:

a^{5}b^{3}

Step-by-step explanation:

The law of indices can be used to simplify mathematical expressions involving  arithmetical operation on variables with powers.

a^{m} x a^{n} = a^{(m+n)}

Thus, the given expression can be simplified as follows:

a³b² a²b  = a³ x a² x b² x b^{1}

               = a^{3+2} x b^{2+1}

               = a^{5}b^{3}

Thus,

a³b² a²b  = a^{5}b^{3}

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To make a sports drink for the football team, Jaylen filled \dfrac9{10}
Scilla [17]

The total gallons of sports drink Jaylen made using 9/10 of a large cooler with water and 6 cups of sports drink concentrate is 3.75 gallons

<h3>Total gallons</h3>

  • Water in the large cooler = 9/10

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= 10-9 / 10

= 1/10

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  • Total gallons of sport drink = x

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Cups to gallons:

16 cups = 1 gallon

Total gallons of sport drink, x = 60 cups / 16

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Complete question:

To make a sports drink for the football team, Jaylen filled 9/10 of a large cooler with water. Then, he filled the remaining space with 6 cups of sports drink concentrate. How many gallons of sports drink did Jaylen make?

Learn more about total gallons:

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stealth61 [152]
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2 years ago
Graph the direct variation equation. <br> Y=-1/2x
Fiesta28 [93]

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3 years ago
The annual energy consumption of the town where Camilla lives in creases at a rate that is onal at any time to the energy consum
boyakko [2]

Answer:

6.575 trillion BTUs

Step-by-step explanation:

<em>Let represent the annual energy consumption of the town as E</em>

<em>The rate of annual energy consumption *  energy consumption at time past</em>

<em>dE/dt * E</em>

<em>dE/dt =K</em>

<em>k = the proportionality constant</em>

<em>c= the integration constant</em>

<em>(dE/dt=) kdt</em>

<em>lnE = kt + c</em>

<em>E(t) = e^kt+c ⇒ e^c e^kt  e^c is a constant, and e^c = E₀</em>

<em>E(t) = E₀ e^kt</em>

<em>The initial consumption of energy is E(0)=4.4TBTU</em>

<em>set t = 0 then</em>

<em>4.4 = E₀ e⇒ E₀ (1) </em>

<em>E₀ = 4.4</em>

<em>E (t) = 4.4e^kt</em>

<em>The consumption after 5 years is t = 5, e(5) = 5.5TBTU</em>

<em>so,</em>

<em>E(5) = 5.5 = 4.4e^k(5)</em>

<em>e^5k = 5/4</em>

<em>We now take the log 5kln = ln(5/4)</em>

<em>5k(1) = ln(5/4)</em>

<em>k = 1/5 ln(5/4) = 0.04463</em>

<em>We find  the town's annual energy consumption, after 9 years</em>

<em>we set t=9  </em>

<em>E(9) = 4.4e^0.04463(9)</em>

<em>= 4.4(1.494301) = 6.5749TBTUs</em>

<em>Therefore the annual energy consumption of the town after 9 years is </em>

<em>= 6.575 trillion BTUs</em>

<em />

3 0
3 years ago
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