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4 years ago

Why are there two solutions for the equation |6 + y| = 2? Explain.

Mathematics

2 answers:

dem82 [27]4 years ago

5 0

Let x = 6+y. We can replace 6+y with x getting the equation |x| = 2.

The equation |x| = 2 has two solutions. The expression |x| represents the distance from x to 0 on the number line.

The two solutions to |x| = 2 are x = 2 and x = -2.

Going from x = 2 to 0 is a distance of 2 units, so is the distance from 0 to x = -2.

The solutions in terms of x are then used to find the solutions in terms of y

For example, if x = 2, then

x = 6+y

2 = 6+y

2-6 = y

y = -4

The other y solution is handled in a similar fashion.

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Side note: The 2 at the end of the original equation does not determine how many solutions there are. We could easily have |6+y| = 3 and still have two solutions.

Drupady [299]4 years ago

5 0

Let x = 6+y. We can replace 6+y with x getting the equation |x| = 2.

The equation |x| = 2 has two solutions. The expression |x| represents the distance from x to 0 on the number line.

The two solutions to |x| = 2 are x = 2 and x = -2.

Going from x = 2 to 0 is a distance of 2 units, so is the distance from 0 to x = -2.

The solutions in terms of x are then used to find the solutions in terms of y

For example, if x = 2, then

x = 6+y

2 = 6+y

2-6 = y

y = -4

The other y solution is handled in a similar fashion.

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Ganezh [65]

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3 years ago

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kvasek [131]

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3 years ago

Match the equation with its corresponding solution for x. 4x + 7 = 35

lyudmila [28]

Are you wanting to solve for x? If so:
4x + 7 = 35
We want to isolate the variable, so subtract 7 from both sides. You will get:
4x = 28
Now, divide each side by 4 to get X by alone. This will give you:
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I hope this helped.

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3 years ago

Find the value of angle "A" and angle "B".

Daniel [21]

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3 years ago

Read 2 more answers

A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria

Kaylis [27]

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$ .

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$ , we get

$\frac{dP}{dt} = kP$ .

Notice that this is a separable equation, so

$\frac{dP}{P} = kdt$ .

Then, integrating in both sides of the equality:

$\int\frac{dP}{P} = \int kdt$ .

We have,

$\ln P = kt+C$ .

Now, taking exponential

$P(t) = Ce^{kt}$ .

The next step is to find the value for the constant $C$ . We do this using the initial condition $P(0)=400$ . Recall that this is the initial population of bacteria. So,