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tatiyna
4 years ago
7

Is the coordinate (1, 2) a solution of the system below?

Mathematics
2 answers:
Stella [2.4K]4 years ago
7 0
Test them
(x,y)
x=1 and y=2

x+2y=5
1+2(2)=5
1+4=5
5=5
true

y=x+1
2=1+1
2=2
true

yes
my answer is A
adell [148]4 years ago
6 0
Hello there!

x + 2y = 5
y = x + 1

We gonna solve y = x + 1 for y
Let's start solving the equation by substitute x + 1 for y in x + 2y = 5
x + 2y = 5 (we gonna replace y by x +1)
x + 2( x + 1) = 5
x + 2x + 2 = 5
3x + 2 = 5
3x = 5 - 2
3x = 3
x = 3/3
x = 1

Since we have the value for x, it will be easier for us to find y. In order to find y we just need to replace x by its value which is 1. So let's go!

We have y = x + 1 >> so we gonna substitute 1 for x
y = x + 1
y= 1+1
y=2
see.... easy :)

The final answer is: (1,2)
The correct option is A (Yes)


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nadya68 [22]

Answer:

1) yes

2) No

3) No

4) yes

5) No

Step-by-step explanation:

1) f(0)=0, f(1)=1, f(2)=2, f(3)=3, then f[\{0, 1, 2, 3\}]=\{0, 1, 2, 3\}

2) f(0)=0, 1^2=1\equiv 1 \text{mod 4}, then f(1)=1; 2^2=4\equiv 0 \text{ mod 4}, then f(2)=0; 3^2=9\equiv 1 \text{mod 4}, then f(3)=1

Then f[\{0, 1, 2, 3\}]=\{0, 1, 3\}, this means that f isn't onto.

3.

  • f(0)=0;
  • 1^1-1=0, then f(1)=0
  • 2^2-2=2, then f(2)=2
  • 3^2-3=6\equiv 2 \text{ mod 4}, then f(3)=2

Then  f[\{0, 1, 2, 3\}]=\{0, 2\}, this means that f isn't onto.

4.  f[\{0, 1, 2, 3\}]=\{0, 1, 2,3\}, then f is onto.

5. f[\{0, 1, 2, 3\}]=\{1, 2\}, this means that f isn't onto.

6 0
3 years ago
Help...* * +++:)&gt; ......
jarptica [38.1K]

Answer:

Hey there!

7(3x-1)-(x+5)=-52

21x-7-x-5=-52

20x-12=-52

20x=-40

x=-2

4 0
3 years ago
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