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iris [78.8K]
3 years ago
12

What is the counterexample for the city of Portland is in Oregon

Mathematics
2 answers:
Karolina [17]3 years ago
6 0
A counterexample<span> is a special kind of </span>example<span> that disproves a statement or proposition. </span>

The counter example for : the city of Portland is in Oregon would be : the city of Portland is in Maine.

The city named Portland can be found in both states.
In-s [12.5K]3 years ago
4 0

One <em><u>possible counterexample</u></em> would be:

The city of Portland is in Connecticut.

Explanation:

A counterexample is something that proves a statement wrong.  Since the city of Portland is in multiple states (Oregon, Maine, Connecticut, Arkansas, etc.), any of the other states having a Portland would serve as a counterexample.

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Find the distance between<br> 2 - 4<br> and<br> 6 + i
saveliy_v [14]

Answer:

Step-by-step explanation:

hello :

let :    Z1 = 2-4i       Z2 = 6+i  

the distance between Z1  and  Z2 is :

/Z2 - Z1 / = /(6+i)-(2-4i)/ =/4+5i/ = √(4²+5²) =√41

3 0
3 years ago
A circle is defined by the equation x2 + y2 + 8x + 22y + 37 = 0. What are the coordinates for the center of the circle and the l
inn [45]
We can say that:
8x = -2ax
8 = -2a
a = -4

22y = - 2by
22 = -2b
b = -11

37 = a² + b² - r²
37 = (-4)² + (-11)² - r²
r² = 16 + 121 - 37
r² = 100
r = 10

With all this information, we can say that:
The center of the circle is: (-4 ; -11)
and the radius of the circle is r = 10

P.S:
(a - x)² + (b-y)² = r²
a² - 2ax + x² + b² -2by + y² - r² = 0
x² + y² -2ax - 2by + a² + b² - r² = 0

6 0
4 years ago
Y intercept of this graph​
Aleksandr [31]

Answer:

The y intercept is - 2

Step-by-step explanation:

Because in the y axis the line crosses - 2

8 0
3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
Of the cakes at Naomi's Bakery, 1/6 have chocolate frosting. Of the cakes with chocolate frosting, 1/5 have raspberry filling. W
Sever21 [200]

Answer:

\frac{1}{6}  \times  \frac{1}{5 }  \\  \frac{1}{30}

5 0
3 years ago
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