Answer:
equation 1
y = -2x
equation 2
y = x-3
Solution to both
(1,-2)
Step-by-step explanation:
We need to get the equations of both lines
General form is;
y = mx + c
where m is slope and c is the y-intercept
Table 1
since we have a point 0,0; the y-intercept here is zero
Let us get the slope. We can do this by selecting any two points
m = (y2-y1)/(x2-x1)
m = (2-10)/(-1+5) = -8/4 = -2
So the equation of the first line is;
y = -2x
Table 2
we get the slope
m = (4+2)/(7-1) = 6/6 = 1
The partial equation is;
y = x + c
To get c, we select any two point and substitute
4 = 7 + c
c = 4-7
c = -3
So the equation is;
y = x-3
To get the solution to both systems, we equate the y
-2x = x - 3
-2x-x = -3
-3x = -3
x = -3/-3
x = 1
To get y, we substitute;
recall; y = -2x
y = -2(1)
y = -2
Solution to the system is;
(1,-2)
Answer:
the lines are neither perpendicular or parallel
Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as
<span>(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). </span>
<span>Next, complete the square. </span>
<span>x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² </span>
<span>Also, for the x in the numerator </span>
<span>x = x + 1/2 - 1/2. </span>
<span>So </span>
<span>(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. </span>
<span>Integrate term by term to get </span>
<span>∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C </span>
<span>b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. </span>
<span>∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C </span>
<span>= 2 √(x) ln(x) - √(x) + C. </span>
<span>c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. </span>
<span>∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C </span>
<span>= arctan(e^x) + C.</span>
Answer:
Step-by-step explanation:
answer to the question:
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The answer would be 5.291 10 to the power of 12
