Which graph represents an exponential function?

Mathematics

1 answer:

schepotkina [342]3 years ago

8 0

Answer:

Presumably the first graph solely based on its shape.

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X = ? y = ? 16 45 degrees

mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}$

For x, we will be using the Cosine Function:

$\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}$ $(16)Cosine(45^{\circ})\text{ = x}$ $(16)(\frac{1}{\sqrt[]{2}})\text{ = x}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = x}$

Therefore, x = 8√2.

For y, we will be using the Sine Function.

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}$ $\text{ (16)Sine }(45^{\circ})\text{ = y}$ $\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = y}$

Therefore, y = 8√2.