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4 years ago

(a^-1+b^-1)/(a^-2-b^-2)Plz help it's urgent

Mathematics

1 answer:

Tasya [4]4 years ago

3 0

Answer: ab/b-a

hope it helps

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Answers: (0,12)<br> (0,0)<br> (cannot be determined) <br> (6,0)<br> Giving brainliest!

Nataly_w [17]

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3 years ago

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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.

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2 years ago

Last year 63 inches of rain fell oaktown. In pine city 7 inches of rain fell last year. Write and solve equation to find how man

Nonamiya [84]

Answer:

Oaktown had 9 times more inches of rain than Pine City did.

Step-by-step explanation:

63 divided by 7 = 9.

Very simple problem. Glad to help.

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4 years ago

(a^-1+b^-1)/(a^-2-b^-2)Plz help it's urgent​

(a^-1+b^-1)/(a^-2-b^-2)Plz help it's urgent