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ollegr [7]
3 years ago
12

How can you tell the difference between systems with no solution and infinite solutions?

Mathematics
2 answers:
antoniya [11.8K]3 years ago
8 0

Answer:

The given system of equations has "No solution” means that the constants are the numbers without variables. If the coefficients are the equal on both sides then their constants will not equal,they did not satisfy the given equations and therefore it said to have no solution

The given system of equations has "infinitely many solutions" if the given linear system of equations has both the equations represents to the same line.

Step-by-step explanation:

natita [175]3 years ago
7 0

No solution would mean that there is no answer to the equation. It is impossible for the equation to be true no matter what value we assign to the variable. Infinite solutions would mean that any value for the variable would make the equation true.

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A population of a particular yeast cell develops with a constant relative growth rate of 0.4225 per hour. The initial population
Ksju [112]

Answer: The population after 3 hours is 13.9 mill.

Step-by-step explanation:

When we have an exponential with:

A  = initial population.

r = constant relative growth rate:

t = time.

The function that models this is:

P(t) = A*e^(k*t)

In this case we know that:

A = 3.9 mill.

r = 0.4225 1.

Then the function that models the population of this yeast cell is:

P(t) = (3.9 mill)*e^(0.4225*t)

where t represents the time in hours.

Then if we want to know the population after 3 hours, we should replace t by 3.

P(3) = (3.9 mill)*e^(0.4225*3) = 13.85 mill.

And we want to round our answer to one decimal place, then we must look at the second decimal place, we can see that is a 5, so we should round up.

The population after 3 hours is 13.9 mill.

8 0
3 years ago
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3 0
3 years ago
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3 years ago
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Step-by-step explanation:

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maw [93]

Answer:

Approximately 12.04 units.

Step-by-step explanation:

The find the distance between any two points, we can use the distance formula, which is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

We have the points (2,7) and (-6,-2). Let's let (2,7) be (x₁, y₁) and let's let (-6, -2) be (x₂, y₂). Substitute:

d=\sqrt{(-6-2)^2+(-2-7)^2

Subtract:

d=\sqrt{(-8)^2+(-9)^2

Square:

d=\sqrt{64+81}

Add:

d=\sqrt{145}

Approximate

d=\sqrt{145}\approx12.04

So, the distance between (2,7) and (-6,-2) is approximately 12.04 units.

And we're done!

7 0
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