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Ronch [10]
3 years ago
7

8) Which of the following is equivalent to 33 + 77

Mathematics
1 answer:
sergiy2304 [10]3 years ago
3 0
The answer is 11(3+7) duh
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Help quick plz i will make you a brainllest
blsea [12.9K]

Answer: B

Step-by-step explanation:

LJ should corresponds with RQ but it should correspond with PR.

8 0
3 years ago
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A highway engineer specifies that a certain section of roadway covering a horizontal distance of 2km should have a downgrade of
Zigmanuir [339]

Answer:

0.16km

Step-by-step explanation:

A highway engineer wants to compute  the change in elevation of a section of road. The horizontal distance of this section of road is 2km and downgrade is 8%

The slope formula is given by

m=\frac{rise}{run}=\frac{x}{y}

m = 8% = 8/100

run = y = 2km

m =\frac{x}{y}

\frac{8}{100} =\frac{x}{2}

\frac{2*8}{100} =x

x=\frac{16}{100} =0.16km

Verification:

m=\frac{0.16}{2}= 0.08=8%%

3 0
3 years ago
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Add -2x^4+x^2-x-9 and x^4-x^3-5x+3
eimsori [14]

Answer:

Your solution would be -1x^4 - 1x^3 + x^2 - 6x - 6.

Step-by-step explanation:

5 0
3 years ago
1. What percent of 25 is 14?
bagirrra123 [75]

Answer:56%

Step-by-step explanation:

8 0
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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