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3 years ago

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Name a possible value for w that is a solution to w > 30, but is not a solution to w < 50. Thank you in advance for the he

lp.

1 answer:

MrMuchimi3 years ago

4 0

Answer:

i dont know

Step-by-step explanation:E

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What is the GCF of 48m5n and 81m2n2?

Lady bird [3.3K]

To find the GCF of the two terms, continuous division must be done.
What can be used to divide both terms such that there is not a remainder?
Start small, let's take 2. It could be a GCF.
Move up higher, say 3. Yes, it can be a GCF.
To see if there might be a greater common factor, divide the constants by 3.
48/3 = 16
81/3 = 27
Upon inspection and contemplation, there is no more common factor between 16 and 27. So, 3 is the GCF.

Moving on, when it comes to variables. The variable with the least exponents is easily the GCF. For the variable m, the GCF is m2 and for n, the GCF is n.

Combining the three, we have the overall GCF = 3m2n

7 0

3 years ago

Read 2 more answers

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

One number is five more than three times another. If their difference is 35, what is the larger number?

miv72 [106K]

Let x and y be the two unknown numbers. If one number, say x,is 5 more than 3 times another, say y, the equation would be

x = 5 + 3y

When their difference is 35, x-y = 35, since x is the much larger number here. We substitute x to solve for the smaller number, y.

x - y = 35
(5 + 3y) -y = 35
5 +2y = 35
2y = 35-5
2y = 30
y = 15

The larger number x would be
x = 5 + 3(15)
x = 50

The answer is 50.

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3 years ago

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Aisha made bracelets for a craft fair. She spent $29.75 on supplies to make the bracelets. Each bracelet cost her about $0.85 to

Alchen [17]

Amount of money Aisha receive from selling all of the bracelets is $33.25.

<u>Step-by-step explanation:</u>

In this question, its given that Aisha made bracelets for a craft fair. She spent $29.75 on supplies to make the bracelets. Each bracelet cost her about $0.85 to make. Aisha sells each bracelet for $0.10 more than it cost her to make. Let's calculate Number of bracelets Aisha made as $\frac{total-cost}{cost-of-one-bracelet}$ .

Number = $\frac{total-cost}{cost-of-one-bracelet}$

⇒ Number = $\frac{total-cost}{cost-of-one-bracelet}$

⇒ Number = $\frac{29.75}{0.85}$

⇒ Number = $35$

Now , its given that Aisha sell these bracelets with $0.10 more then it cost her i.e. at $0.95 , Let's calculate selling price as :

Price = $0.95(35)$

Price = $33.25

Therefore, amount of money Aisha receive from selling all of the bracelets is $33.25.

4 0

4 years ago

What are the solutions to x2 – 10x = 39?

oee [108]

Answer: x=−3 or x=13

Step-by-step explanation:

Step 1: Subtract 39 from both sides.

x2−10x−39=39−39

x2−10x−39=0

Step 2: Factor left side of equation.

(x+3)(x−13)=0

Step 3: Set factors equal to 0.

x+3=0 or x−13=0

x=−3 or x=13

7 0

2 years ago