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Artemon [7]
4 years ago
13

This is 70 points I will report anyone who answers wrong just for points! I will be asking TWO questions Please answer correctly

or I will report you :)
question 1 : (x+3)y=14 which ordered pair in the form (x,y) is a solution of this equation
(11, 1)
(3, 2)
(5, 2 ) (7, 2)

second question: which pairs are solutions to the equation
5xy+9=44

1, 7 and 4, 11
1, 7 and 7, 1
7,1 and 2, 3
4, 11 and 2, 3
Mathematics
1 answer:
Ratling [72]4 years ago
8 0




QUESTION 1: In these type of question, the easiest way to get the answer is try to plug in the x and y values from the options given in the equation given, So in the first question all the choice except C are more then 14 if you plug in x and y's, for eg, if you plug in x = 3 and y = 2 , you get (3+3)2 = 14 6 x 2 = 14 12 is not equal to 14, so this eliminates this choice but if you chose C you get, (11+3)1 = 14 14 = 14 so this makes C the solution for first question and for the second question do the same thing, and the answer will be D. Hope this helps




QUESTION 2: 5xy + 9 = 44

5xy = 35

xy = 7

solution pairs are:

C. (1, 7) and (7, 1)

not mentioned: (-1.-7) and (-7, -1)






Hope this helps



Plzz don't forget to rate and thanks me


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Answer:  

The smallest value of p+q is 11

It happens when p = 6 and q = 5.

=======================================================

Explanation:

Let's factor 180 in such a way that exactly one factor is a perfect square.

I'll ignore the trivial factor of 1.

Here are the possible factorizations we could go with:

180 = 4*45

180 = 9*20

180 = 36*5

Those factorizations then lead to the following

\sqrt{180} = \sqrt{4*45} = \sqrt{4}*\sqrt{45}= 2\sqrt{45}\\\\\sqrt{180} = \sqrt{9*20} = \sqrt{9}*\sqrt{20}= 3\sqrt{20}\\\\\sqrt{180} = \sqrt{36*5} = \sqrt{36}*\sqrt{5}= 6\sqrt{5}\\\\

Then we have

p+q = 2+45 = 47

p+q = 3+20 = 23

p+q = 6+5 = 11

The smallest value of p+q is 11 and it happens when p = 6 and q = 5.

Side note: p+q is smallest when we go with the largest perfect square factor.

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2 years ago
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I:2x – y + z = 7
II:x + 2y – 5z = -1
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you can first use III and substitute x or y to eliminate it in I and II (in this case x):
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I': 2*(6+y)-y+z=7
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y+z=-5

II':(6+y)+2y-5z=-1
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then you can subtract II' from 3*I' to eliminate y:
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