All categories

3 years ago

How many eight-sized parts do u need to model 3/4?explain

Mathematics

1 answer:

Oliga [24]3 years ago

6 0

Answer:

Step-by-step explanation:

$\dfrac{3}{4}=\dfrac{3}{4}\cdot\dfrac{2}{2}\\\\=\dfrac{3\cdot 2}{4\cdot 2}=\dfrac{6}{8}$

3 fourths = 6 eighths

It will take 6 eighth-sized parts to model 3/4.

You might be interested in

Find the value of x. F belongs to DG PLEASE NEED ANSWER ASAP!

postnew [5]

Answer:

x = 34

Step-by-step explanation:

F = 180 - (90 + 42)

F = 48

2x + 64 + 48 = 180

2x = 180 - 64 - 48

2x = 68

x = 68/2

x = 34

4 0

3 years ago

Hello pls help meeee

Andrej [43]

Answer: 96

You add 12 every week, so you can multiply 12 by 8 (12 x 8) and you get 96!

4 0

3 years ago

Read 2 more answers

What is the coefficient in the expression 10x2-7?

NeX [460]

Answer:

10 is the coefficient of x

7 0

3 years ago

Read 2 more answers

What is the equation of the following line?

irina1246 [14]

E.y=1/4x
if you plug the points into slope-intercept form, you can find the slope.

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 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

4 years ago