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Galina-37 [17]
4 years ago
5

Need help! I have no clue how to even go about solving this..

Mathematics
1 answer:
Solnce55 [7]4 years ago
6 0

In Problem 13, we see the graph beginning just after x = -2. There's no dot at x = -2, which means that the domain does not include x = -2. Following the graph to the right, we end up at x = 8 and see that the graph does include a dot at this end point. Thus, the domain includes x = 8. More generally, the domain here is (-2, 8]. Note how this domain describes the input values for which we have a graph. (Very important.)

The smallest y-value shown in the graph is -6. There's no upper limit to y. Thus, the range is [-6, infinity).

Problem 14

Notice that the graph does not touch either the x- or the y-axis, but that there is a graph in both quadrants I and II representing this function. Thus, the domain is (-infinity, 0) ∪ (0, infinity).

There is no graph below the x-axis, and the graph does not touch that axis. Therefore, the range is (0, infinity).

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The lengrh of a rectangle is represented by the function L(x) = 5x. The width of that same rectangle is represented by the funct
MrRissso [65]

Step-by-step explanation:

just multiply the 2 together to get the area

it would just be

(5x) \times (2{x}^{2} - 4x + 13)

use the distributive property

10 {x}^{3}  - 20 {x}^{2}  + 65

unless you want to simplify the answer, this would be the answer

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5(2{x}^{3}  - 4 {x}^{2}  + 13)

7 0
2 years ago
Consider points A(1, 6) and B(8, 8). Find point C on the x-axis so AC +BC is a minimum.
Kay [80]

Answer:

The coordinates of the point C that minimizes AC + BC are (-20, 0) or (4, 0)

Step-by-step explanation:

The given coordinates of the points A and B are A(1, 6) and B(8, 8)

The location of the point C = The x-axis

Therefore;

The coordinates of the point C = (x, 0)

The length of the segment AC = √((1 - x)² + (6 - 0)²) = √((1 - x)² + 6²)

The length of the segment BC = √((8 - x)² + (8 - 0)²) = √((8 - x)² + 8²)

At minimum distance of AC + BC, we have;

d(√((1 - x)² + 6²) +√((8 - x)² + 8²))/dx = 0 = (1 - x) × 2 × (0.5 - 1)× (√((1 - x)² + 6²)^(0.5 - 1) + (8 - x) × 2 × (0.5 - 1)× √((8 - x)² + 8²)^(0.5 - 1)

∴ d(√((1 - x)² + 6²) +√((8 - x)² + 8²))/dx = 0 = -(1 - x)/√((1 - x)² + 6²) - (8 - x)/√((8 - x)² + 8²)

-(1 - x)/√((1 - x)² + 6²) = (8 - x)/√((8 - x)² + 8²)

(8 - x)·√((1 - x)² + 6²) = -(1 - x)·√((8 - x)² + 8²)

Squaring both sides gives;

(8 - x)²·((1 - x)² + 6²) = (1 - x)²·((8 - x)² + 8²)

Expanding, using an online tool, we get;

x⁴ - 18·x³ + 133·x² -720·x + 2368 = x⁴ - 18·x³ + 161·x² - 272·x + 128

Which gives;

(161 - 133)·x² - (272 - 720)·x + 128 - 2368 = 28·x² + 448·x - 2240 = 0

Dividing by 28 gives;

x² + 16·x - 80 = 0

(x + 20)·(x - 4) = 0

Therefore, x = -20 or x = 4

The coordinates of the point C that minimizes AC + BC are (-20, 0) or (4, 0)

4 0
3 years ago
If x to the 2nd power =20 what is the value of x
11111nata11111 [884]
X= 4.47213595 Is the answer to the question just round to the nearest 10th
7 0
3 years ago
Read 2 more answers
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