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Cloud [144]
3 years ago
13

Is there a such thing as a zillion?

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:

yes

Step-by-step explanation:


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11 x 4.5<br> help asap!!!
FromTheMoon [43]

Answer:

49.5

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Solve for n.<br> 0 = lu19
mariarad [96]

Answer:

There is no n in the equation but i think its supposed to be 0=In19 so your answer is: n= 0

Step-by-step explanation:

You can tell off the bat the answer is 0, this is why:

0=In19

make it into a distributive problem so you can get n away from the rest:

0=n (I19)

divide I19 on both sides of the equation:

0/I19 =n

Anything divided or multiplied by 0 is 0:

0=n

8 0
3 years ago
If two angles are supplementary and one of the angles measures 62°, what is the measure of the other angle?
Rzqust [24]

Answer:

118°

Step-by-step explanation:

Supplementary angles add up to 180°.

Since the two angles are supplementary then the two angles add up to 180°. One angle measure is 62°, the other has to be 180° - 62 = 118°

7 0
4 years ago
The graph shown is a translation of the graph of f(x) = x². Write the function in
Sidana [21]

Answer:

plz insert the graph, ill revise my answer

Step-by-step explanation:

6 0
3 years ago
the air in a small room 12ft by 8ft by 8ft is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is bl
Irina-Kira [14]

Answer:

The air in the room at 0.01% carbon monoxide at 43.8 min

Step-by-step explanation:

Let be the volume of CO in the room at time t, be v(t)  and  the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:

p(t) = \frac{100\times v(t)}{V}

Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.

Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.

As you wrote, the rate at which the volume of CO in the room changes with time is given by

\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)

This is a simple first-order equation:

\frac{dv}{v} = -\frac{f}{v} dt

ln(v) - ln(c) = -\frac{f}{v} \times t

where ln(c) is the constant of integration.

ln \frac{v}{c} = -\frac{f}{v} \times t

v(t) = c \times e^{(-f*\frac{t}{V})}

In terms of volume percent,

p(t) = \frac{100*v(t)}{V}= (\frac{C}{V})*exp(\frac{-f \times t}{v})

where C = 100*\frac{c}{V} is just another way of writing the constant.

Plugging in the values for the constants, we get:

p(t) = (\frac{C}{768 cu.ft.})* exp(\frac{-t}{7.68 min})

Now use the initial condition (p(0) = 3% at t = 0) to solve for C:

3% = C

p(t) = (3\%)\times exp(\frac{t}{7.68 min})

To find the time when the air in the room reaches a certain value, it is easier to rewrite this solution as:

\frac{p(t)}{3\%} = exp(\frac{-t}{7.68 min})

t(p) = -(7.68 min)ln(\frac{p}{3\%} )

= (7.68 min)*ln(\frac{3\%}{p})

The question asks when p(t) = 0.01%. Plugging this into the above equation, we get:

t(0.01\%) = (7.68 min)*ln(\frac{3}{0.01}) = 43.8 min

3 0
3 years ago
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