Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin

3 0

3 years ago

Graph the line that passes through the points (7, 0) and (-2, 9) and determine the equation of the line.

MariettaO [177]

Answer:

y = -1x + 7

Step-by-step explanation:

y2 - y1 / x2 - x1

9 - 0 / -2 - 7

9 / -9

= -1

y = -1x + b

0 = -1(7) + b

0 = -7 + b

7 = b

5 0

2 years ago

N^2+6n+c<br><br> Find the value of c that completes the square

Anon25 [30]

Answer:

The value of ”c” can by found by this formula: (b/2)^2. In this equation the ”b” is 6. So, (6/3)^2=3^2=9.

The value of “c” that completes the square is 9.

6 0

2 years ago

Graph the set. in number line<br> (1, ∞) ∩ (−10, ∞)

svlad2 [7]

See attachment for the graph of the set (1, ∞) ∩ (−10, ∞)

<h3>How to graph the set on a number line?</h3>

The set is given as:

(1, ∞) ∩ (−10, ∞)

The above means that the numbers inclusive in the set are numbers between the following set:

(1, ∞): All real numbers greater than 1
(−10, ∞): All real numbers greater than -10

The intersection of the above set is (1, ∞)

This means that the set (1, ∞) ∩ (−10, ∞) is a set of all real numbers greater than 1

Having done the above illustration, next we plot the set on a number line

See attachment for the graph of the set (1, ∞) ∩ (−10, ∞)