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3 years ago

What if the following is the sum of both solutions of the equation....

Mathematics

1 answer:

LuckyWell [14K]3 years ago

6 0

answer

D 2

factor

factor the equation $x^{2} - 2x - 8 = 0$ into (x + a) and (x + b) so that a + b = -2 and a * b = -8

the two numbers that meet these conditions are a = -4 and b = 2

$x^{2} - 2x - 8 = (x - 4)(x + 2)$

find solutions

set both (x - 4) and (x + 2) equal to zero to find your solutions

x - 4 = 0

x = 4

x + 2 = 0

x = -2

add solutions

since the two solutions are x = 4 and x = -2, add them together to get your final answer

4 + (-2) = 2

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Answer:
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Answer:

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Step-by-step explanation:

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A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain

AysviL [449]

Answer:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

Step-by-step explanation:

For this case we have the following value:

$R = 0.0498 \frac{Kpa}{Km}$

We can convert this first to $\frac{Kpa}{m}$ like this:

$R=0.0498 \frac{Kpa}{Km} *\frac{1km}{1000m}=0.0000498 \frac{Kpa}{m}$

Now we use the fact the the pressure is defined as $P =\frac{F}{A}$ , whre P is the pressure, F the force and A the area, so then $Kpa= \frac{KN}{m^2}$ and then we can replace this:

$R=0.0000498 \frac{KN}{m^3}$

Now from definition of work we know that $W= Fd$ where W is the work, F the force and d the distance, so then is equivalent $KN =\frac{KJ}{m}$

And if we replace this into the equation we got:

$R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}$

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