Which expression is equivalent to 5m + m + 2m?

Saul earned $25 for painting his neighbor's fence. He spent $11 at the store and put the rest into his bank account.

PtichkaEL [24]

The answer is $14 Saul put into his bank accout

3 0

2 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using reductio ad absurdum. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by reductio ad absurdum that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

3 years ago

Please help me with this 20 points

JulsSmile [24]

Answer:

A 4/605

Step-by-step explanation:

mathematics = 11 letters

P(m) = number of m's/total = 2/11

we replace it so we have mathematics = 11 letters

P(s) = number of s's/total = 1/11

not replacing it

we replace it so we have mathematic = 10 letters

P(vowel) = number of vowel/total = 4/10 = 2/5

P(m, replace,s, no replace, vowel) = 2/11* 1/11 * 2/5 = 4/605

3 0

3 years ago

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What’s the answer and explain it please -

MariettaO [177]

Answer:

2.6

Step-by-step explanation:

The Mean Absolute Deviation (MAD) is the average amount that each point is from the overall average of the data set.

So, first find the overall average (also called the mean):

avg = (1 + 1 + 2 + 3 + 5 + 7 + 9) / 7

= 28 / 7

= 4

The average (mean) is 4, so now you calculate how far each point is from 4 and average those numbers:

Note: Remember to use absolute values, since distance can't be negative.

4 - 1 = 3

4 - 2 = 2

4 - 3 = 1

4 - 5 = |-1| = 1

4 - 7 = |-3| = 3

4 - 9 = |-5| = 5

And finally, average (or find the mean) the new data set:

avg = (1 + 1 + 2 + 3 + 3 + 3 + 5) / 7

so 18 / 7 is the MAD of the data set.

Finally, round it to the nearest tenth:

18 ÷ 7 = 2.57142857143 ≈ 2.6

7 0

2 years ago

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7 L of an acid solution was mixed with 3 L of a 15% acid solution to make a 29% acid solution. What is the percent concentration

Setler [38]

well, let's say that the first solution is 7 liters and has "x" percent of acid, how much acid total in it? well, 7x, keeping in mind that "x" is a decimal form.

likewise, the second solution is 3 liters and has 15% acid, so how much acid is there in that one? (15/100) * 3 = 0.45, so

$\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&7&x&7x\\ \textit{2nd solution}&3&0.15&0.45\\ \cline{2-4}&\\ mixture&10&0.29&2.9 \end{array}~\hfill \implies 7x~~ + ~~0.45~~ = ~~2.9 \\\\[-0.35em] ~\dotfill\\\\ 7x=2.45\implies x=\cfrac{2.45}{7}\implies x=0.35~\hfill \stackrel{\textit{converting to \%}}{0.35\cdot 100}\implies \stackrel{\%}{35}$

5 0

2 years ago

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