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umka21 [38]
4 years ago
8

What value of n solves the equation 3n=1/81 N=___

Mathematics
1 answer:
Radda [10]4 years ago
7 0

3n = 1/81; divide both sides by 3 to isolate n:


3n/3 = 1/81 divided by 3. Remember that when you divide, you use the reciprocal and multiply instead!!


n = 1/81(1/3) = 1/243. Now plug 1/243 back in for n, and 3/243 is equivalent to 1/81

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If 25x2- 49y2 has one factor (5x-7y), then the other factor is…..
elena-s [515]
There's only one 1 factor and there are no like terms
6 0
3 years ago
May somebody help me please.
MakcuM [25]

The answer is four!!

8 0
3 years ago
Plot C(-1,–7), D (6,–7), E(5,3), and F(5,-5) in a coordinate plane. Are CD and EF congruent?
N76 [4]

Answer:

No, they are not congruent

Step-by-step explanation:

By CD and EF being congruent, we want to know if they are equal.

What we can simply do here is to find the distance between the points present.

That is, the distance between C and D, the distance between E and F

To get the distances between two points , we use the following equation

D = √(x2-x1)^2 + (y2-y1)^2

For CD, we have

D = √(6 -(-1)^2 + (-7 + 7)^2

D = √(7)^2 = 7 units

For EF, we have

D = √(5-5)^2 + (-5-3)^2

D = √(-8)^2

D = √(64) = 8 units

This clearly shows that they are both not congruent

7 0
4 years ago
How would you convert 93/20 to a decimal?
anastassius [24]
In order for you to find the decimal for this problem you would have to divide both of  the numbers from each other.

So for this problem you have to 

Problem → \frac{93}{20}

Write them out  → 93 ÷ 20 

Answer → 4.65

So, that means that your answer is 4.65


4 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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