Ask question

Ask question

All categories

SVETLANKA909090 [29]

3 years ago

9

Deb has of a brownie. She wants to divide it up into 's of a brownie so she can have it throughout the day. How many 's will she

have?

1 answer:

qaws [65]3 years ago

4 0

The answer would be 3

You might be interested in

The sign of the product of two integers with the same sign as positive. What is the sign of the product of the three inegers wit

vladimir1956 [14]

Answer:

<h2>The same sign as the original numbers</h2>

Step-by-step explanation:

The best way to explain this is show it in action.

EXAMPLE 1:

-3 × -6 × -17 = -306 (Same sign)

EXAMPLE 2:

6 × 4 × 9 = 216 (Same sign)

I'm always happy to help :)

8 0

3 years ago

Riley has 24 more stickers than Maria.

never [62]

Answer:

n+24=x

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

How do I find the gradient of the line y=x-5

Katen [24]

The gradient is the same as the slope.
The gradient is always before the x variable or is the coefficient of the x variable when an equation is in the slope intercept form.
The gradient in this equation is 1

8 0

2 years ago

How many solutions exist for the given equation?

bogdanovich [222]

Answer: Infinitely many solutions when we graph this it comes out as one straight line also try using Desmos it helps with equations like this by graphing them for you! -Your friend, Bill Cipher

Step-by-step explanation: Have a great Valentines day <3

4 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