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guajiro [1.7K]
3 years ago
5

PLEASE HELP!

Mathematics
1 answer:
statuscvo [17]3 years ago
5 0

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Dimensions of the photo is given by

20\ in.\ by\ 24\ in.

According to question, the photo is reduced so that the longer dimension is 15 inch.

So, Let the width of the reduced photo be x.

So, it becomes,

\frac{20}{x}=\frac{24}{15}\\\\x=\frac{15\times 20}{24}\\\\x=12.5\ in.

So, the width of the reduced photo is 12.5 inch.

Hence, Option 'B' is correct.

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Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

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From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

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Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

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Step-by-step explanation:

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