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3 years ago

Find the value of the variable.

Mathematics

2 answers:

UNO [17]3 years ago

7 0

Answer:

x=8

base side of isosceles triangle

y=√{8²+8²}=8√2

Zielflug [23.3K]3 years ago

3 0

Answer:

x = 8, y = $8\sqrt{2}$

Step-by-step explanation:

$a^2+b^2=c^2$

$8^2+8^2=c^2$ Both 8 and x sides are congruent so they're equal.

$64+64=c^2$ Add 64 + 64.

$c^2=128$ Square root 128

$c^2=8\sqrt{2}$

Final answers: x = 8, y = $8\sqrt{2}$

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2 years ago

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Answer:

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3 years ago

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2 years ago

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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2 years ago