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inysia [295]
2 years ago
8

Suppose the total rent for the apartment is increased by $150. How much would Sam's share be then??? Help please

Mathematics
1 answer:
lys-0071 [83]2 years ago
7 0
$150 divided amongst himself and friends (3) which equals 50, so add $50 to Sam’s share.
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What is the Absolute value of -6
DiKsa [7]

Answer:6

Step-by-step explanation:

5 0
3 years ago
Please help its my last question and I really need to submit this right about now. I’m not sure what the answer is and how I do
iren [92.7K]

Answer:

I legit answered this its 6/10

Edit: y = -6/10x + 100

Step-by-step explanation:

Change in y over change in x

60/100

6/10

8 0
3 years ago
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The ratio of the number of cars to the number of motorcycles in a parking lot is 10 : 3. The ratio of the number of motorcycles
Harman [31]
5 because 20:6 is the car to motorcycle. 6:15 is the motorcycle to van. 20÷15=5 ~JZ
3 0
3 years ago
What is the cosecant of angle C?
pishuonlain [190]
Cosecant = hypotenuse / opposite side    ( that is  1  / sine)
csc C =  AC / 8

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5 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
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