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Sidana [21]
3 years ago
12

Please help, show work! Limits and functions! 85 points!

Mathematics
1 answer:
swat323 years ago
3 0

Answer:

Ok I might misunderstand this but this is what I got ( in order )

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Use the Side-Splitter Theorem to find x, given that PQ||BC . The figure is not drawn to scale.
MrRa [10]

Answer: x=12


Step-by-step explanation:

1. Accoding to the Side-Splitter Theorem, if a line interesect two sides of the triangle and it is parallel to the third side, then the other two sides are divided by the line proportionally.

2. Then, you can form the following proportion:

\frac{PA}{BP}=\frac{QA}{CQ}

3. Substitute values and solve for x:

\frac{8}{x}=\frac{12}{18}\\18*8=12x\\x=12


8 0
3 years ago
How can you find the equation?
katrin2010 [14]
The equation is 2x (or x multiplied by 2) because if you divided 32/16 or 20/10 and so on, all answers will give you the number 2
7 0
3 years ago
Read 2 more answers
Solve 7 ÷ one half = ___.<br> Question 1 options:<br><br> 14<br><br> 15<br><br> 24<br><br> 40
Norma-Jean [14]

Answer:

O 14

Step-by-step explanation:

hope it is helpful :)

7 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
4 years ago
|-4y+2z|-7z z=-9 and y=2
Vesnalui [34]

Answer:

 37 sorry if i calculated wrong

step-by-step explanation:

5 0
3 years ago
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