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Liono4ka [1.6K]

4 years ago

10

Three apples plus twice as many pears add up to one-half the number of grapes in fruit basket. How many grapes are in the basket

?

1 answer:

Anna35 [415]4 years ago

5 0

Answer:

the answer is 18 because 3 apples plus 6 pears 3*2 equal 9 which is have of the so 9*2 equals 18

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A polygon on a coordinate grid was transformed using (x,y) (x-4,y-3 whats the transformation

pantera1 [17]

Answer:

A. A translation 4 units to the left and 3 units down.

Step-by-step explanation:

6 0

2 years ago

Please help, thanks!

Mamont248 [21]

((-4x^3)(y^4))^-3
--------------------------
(2xy^4)^-4
16x^4y^16
=------------------------
-64x^9y^12
-16y^4
=------------------------
64x^5
-y
=------------------------
4x^5

So the final answer is:

-y^4
-------
4x^5

(Btw if "^" is in front of a number, that means it is an exponent, so when your writing this on paper, just write it as a regular exponent without the "^". I had to do that since I'm on a computer.)

Hopefully this was helpful in some way.

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4 years ago

A pot holds 324 milliliters of water. How many liters will 19 pots hold?

Otrada [13]

For the pot one 6,156 litters

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3 years ago

What does each part of an exponential function represent?*<br> f(x) = a(b)* + k

Sergeeva-Olga [200]

A asymptote
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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

4 years ago