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3 years ago

How do you solve the following for x?

Mathematics

2 answers:

EleoNora [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Is called a letter that stands for the number

Pachacha [2.7K]3 years ago

5 0

Answer:

Step-by-step explanation:

√10x - 1 = 7

Square both sides,

10x - 1 = 7²

10x - 1 = 49

10x = 49 + 1

10x = 50

x = 50/10

x=5

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The qualitative counterpart of validity and reliability in quantitative research, which includes credibility, transferability, a

serious [3.7K]

Answer:

Trustworthiness

Step-by-step explanation:

For qualitative researchers, we have

1. Credibility

2. Transferability

3. Dependability

4. Confirmability and

5. Trustworthiness

While for quantitative researchers, we have

1. Internal validity

2. External validity

3. Reliability

4. Objectivity.

Trustworthiness is the qualitative research counterpart of validity and reliability.

4 0

3 years ago

Iris has 120 songs on her playlist. Each week, she downloads 4 new songs.

aksik [14]

Answer:

4w+120.

Step-by-step explanation:

She downloads 4 songs for every week which is represented as 4w, and she started with 120 songs which is represented as +120.

6 0

2 years ago

Read 2 more answers

-8)(56)=1 What is the missing value that makes the equation true?

Rudik [331]

the correct answer is -448

7 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago

Simplify 26-18b+11(b-3)

xxMikexx [17]

Answer:

-7b-7

Step-by-step explanation:

First you have to distribute 11 throughout the parentheses.

$26-18b+11(b-3)\\26-18b+11b-33$

Now we can combine like terms. (-18b, 11b, -33, 26)

$-18b+ 11b$ and $-33+26$

$-7b$ $-7$

This is your simplified answer for your expression.

-7b-7

6 0

2 years ago

Read 2 more answers