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3 years ago

11

A temperature record in Antarctica was -120.the temperature recorded in the Sahara desert was 129.how many degrees warmer is 129

than -120

1 answer:

DanielleElmas [232]3 years ago

7 0

This is 129 - (-120)

= 129 + 120

= 249 degrees warmer

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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

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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>

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<em>we set t=9 </em>

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<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>

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A farmer produces both corn and wheat. if the price of corn increases, the farmer will supply _____ wheat.

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harkovskaia [24]

Answer:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

Step-by-step explanation:

Given matrix $M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

The Inverse of a 2X2 matrix

$A=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)$

can be found using the following:

$A^{-1}=\dfrac{1}{ad-bc} \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)$

Therefore:

$M^{-1}=\dfrac{1}{(5*-5)-(3*-8)} \left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=-1\left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

Next, we find the product $M^{-1}M$

$M^{-1}M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\\=\left(\begin{array}{ccc}-5*-5+3*-8&-5*3+3*5\\-8*-5+5*-8&-8*3+5*5\end{array}\right)\\=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

Therefore:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

NOTE: The multiplication of a matrix and its inverse always gives the identity matrix as seen above,

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3 years ago