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Vinil7 [7]
3 years ago
9

Pls help simple stuff if you know it. I obviously don't

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

The answer to your question is    y = 2.^{x} + 5

Step-by-step explanation:

Because of the distribution of the points, we conclude that it is an exponential function so we discard the third option (it's a linear function) and the fourth option (it's a quadratic function)

- To solve this problem, evaluate functions 1 and 2 using the x-coordinates and compare the results with the graph

                       y = 2.^{x} + 5                              y = 4.^{x} + 8

1                       y = 2 + 5 = 7                           y = 12

3                      y = 8 + 7 = 15                         y = 21 + 8 = 29

5                      y = 32 + 7 = 39                      y = 1024 + 8 = 1032

7                      y = 128 + 7 = 135                    y = 16384 + 8 = 16392

9                      y = 512 + 7 = 512                    y = 262144 + 8 = 262152

11                      y = 2048 + 7 = 2055             y = 4194312 + 8 = 4194320

13                     y = 8192 + 7 = 8199                y = 67108864 + 8 = 67108872

15                     y = 32768 + 7 =  32775         y = 1073741832 + 8 = 1073741840

Comparing the result with the graph, we conclude that the answer is the first choice.

   

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Find the equation of a straight line passing throught the point listed and having the given gradient. Express your answer in the
Igoryamba
First let's try to find the equation in this form : <span>y = mx + c 

The gradient is given 3 . In a line's equation, x's coefficient represents the line's gradient.

So equation of a line with the gradient of 3, would look like this ;

</span>y= 3x + c
<span>
Now a point that the line passes through is given, (1, 2) 

This point's x-coordinate is 1 and y-coordinate is 2.

So we'll plug its x-coordinate value in the equation and also y-coordinate value. So we can solve it.

As you know, </span>x=1 and y=2

y = 3x + c

2\quad =\quad 3\cdot 1+c\\ \\ 2\quad =\quad 3+c\\ \\ 2-3\quad =\quad c\\ \\ -1\quad =\quad c

We found c = -1 

Also in a line's equation, c is constant and it represents the line's y-intercept

So let's build the line's equation.

m=3 and c=-1

y= mx + c

y= 3x -1

We found the line's equation in this form, y= mx + c

Now let's turn it into this form, ax + by + c = 0

y\quad =\quad 3x-1\\ \\ y-3x\quad =\quad -1\\ \\ y-3x+1\quad =\quad 0\\ \\ -3x+y+1\quad =\quad 0

Final answers,

\boxed { y\quad =\quad 3x-1 }

and 

\boxed { -3x+y+1\quad =\quad 0 }

I hope this was clear enough :)




<span>


</span>
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