R is less than or equal to -9

What is the value of 28% of 95?

Bezzdna [24]

It would be 26.6 as the answer

8 0

3 years ago

17. Find the height of a rectangular room that

vichka [17]

Answer:

the height: h=2,25

Step-by-step explanation:

volume of rectangular box = wide*long*the height

----> the height = volume of rectangular box / ( wide*long) = 2,25

give me 5 stars and hearts^^

4 0

1 year ago

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Find the GCF <img src="https://tex.z-dn.net/?f=32xy-18%20x%5E%7B2%7D%20" id="TexFormula1" title="32xy-18 x^{2} " alt="3

eduard

The GCF is the largest number that can go into both the terms. Since "y" isn't in both terms, it can't include y. x, however, is, so we know our GCF has x in it. What's the largest whole number that can go into 32 and 18? 2. Your GCF is 2x.

Don't forget to rate this answer the Brainliest!

4 0

3 years ago

Solve for x. 3x + 5 = 23

Lunna [17]

[|] Answer [|]

$\boxed{X \ = \ 6}$

[|] Explanation [|]

3x + 5 = 23

_________

Subtract 5 From Both Sides:

3x + 5 - 5 = 23 - 5

Simplify:

3x = 18

Divide Both Sides By 3:

$\frac{3x}{3} \ = \ \frac{18}{3}$

Simplify:

X = 6

_________

- Check Your Work -

Substitute 6 For X:

3 * 6 + 5 = 23

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

3 * 6 = 18

18 + 5 = 23

$\boxed{[|] \ Eclipsed \ [|]}$

8 0

3 years ago

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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:

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

4 0

3 years ago