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3 years ago

What is the range of the function f(x)=-3x-5 with a domain of {-3,0,11}

Mathematics

1 answer:

Sloan [31]3 years ago

5 0

Answer:

R = {4,-5,-38}

Step-by-step explanation:

We know that the domain is the x value, meaning that the range is the y value.

f(x) = -3x - 5 is also y = -3x - 5

ur To find y(range) we need to solve the equation. Since we already have our x-values of {-3,0,11} all we need to do is plug in the values for x and solve.

I think you know the process of how to for y if you plug in these domain numbers, but if you don't, here is how to do it.

y = -3(-3) - 5 y = -3(0) - 5 y = -3(11) - 5

y = 9 - 5 y = 0 - 5 y = -33 - 5

y = 4 y = -5 y = -38

Hope this helps!

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3 years ago

Read 2 more answers

Hi guys

Alekssandra [29.7K]

(5x) / (2/11) = 3 / (1/5)
cross multiply
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3 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)

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3 years ago