You are asked to do this problem by graphing, which would be hard to do over the Internet unless you can do your drawing on paper and share the resulting image by uploading it to Brainly.
If this were homework or a test, you'd get full credit only if you follow the directions given.
If <span>The points(0,2) and (4,-10) lie on the same line, their slope is m = (2+10)/(-4), or m =-3. Thus, the equation of this line is y-2 = -3x, or y = -3x + 2.
If </span><span>points (-5,-3) and (2,11) lie on another line, the slope of this line is:
m = 14/7 = 2. Thus, the equation of the line is y-11 = 2(x-2), or y = 11+2x -4, or y = 2x + 7.
Where do the 2 lines intersect? Set the 2 equations equal to one another and solve for x:
</span>y = -3x + 2 = y = 2x + 7. Then 5x = 5, and x=1.
Subst. 1 for x in y = 2x + 7, we get y = 2(1) + 7 = 9.
That results in the point of intersection (2,9).
Doing this problem by graphing, on a calculator, produces a different result: (-1,5), which matches D.
I'd suggest you find and graph both lines yourself to verify this. If you want, see whether you can find the mistake in my calculations.
Answer:
D
Step-by-step explanation:
The augmented matrix for the system of three equaitons is

Multiply the first row by 5, the second row by -3 and add these two rows:

Subtract the third row from the second:

Divide the third row by 6:

Now multiply the third equation by 26 and add it to the second row:

You get the system of three equations:

From the third equation

Substitute z=2 into the second equation:

Now substitute z=2 and y=5 into the first equation:

The solution is (1,5,2)
(3m²n⁴)³ = 27m^6n^12. choice C
I have no idea what you're talking about. Can you explain in detail plz?