QUICK HELP PLEASE WILL GIVE BRAINLIEST

Almost of the mass of the human body is made up of just six elements.

Vesnalui [34]

Answer:

Step-by-step explanation:

To solve each of these all you have to do is multiply 60kg which is the whole mass of a human body times the percentage so..

Oxygen = (0.65)(60) = 39kg

Carbon = (0.18)(60) = 10.8kg

Hydrogen = (0.10)(60) = 6

Nitrogen = (0.03)(60) = 1.8kg

Calcium = (0.015)(60) = 0.9kg

Phosphorus = (0.01)(60) = 0.6kg

These numbers dont exactly add up to 60 because the percentages gives dont add up to 100%, but this is how to do it.

4 0

3 years ago

Verify that triangle ABC is equal to DEF . Find the scale factor of triangle ABC to triangle DEF

alexira [117]

Answer:

k = 6 : 5 : 4

Step-by-step explanation:

BC : AB : AC = 18 : 15 : 12 = 6 : 5 : 4

EF : DE : DF = 12 : 10 : 8 = 6 : 5 : 4

5 0

2 years ago

Solve 12x-5>-2 please

Stells [14]

Answer:

$x > \frac{1}{4}$

Step-by-step explanation:

$12x - 5 > - 2$

Lets add +5 in both the sides.

$= > 12x - 5 + 5 > - 2 + 5$

$= > 12x > 3$

Now lets divide 12 on both the sides.

$= > \frac{12x}{12} > \frac{3}{12}$

$= > x > \frac{1}{4}$

8 0

3 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 <em>reductio ad absurdum</em>. 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 <em>reductio ad absurdum</em> 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

Which of the following points is in the solution set of y > -x2 + 5?

Stella [2.4K]

Its the third option (2, 4)

7 0

3 years ago

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