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1 year ago

Select the correct answer.

Mathematics

1 answer:

jek_recluse [69]1 year ago

6 0

The value of equation which is given is 12.

According to the Statement

we have given the expression and the some values and we have to find the values of the expression after put the given values in it.

So,The given values are:

h = 8. j = -1, and k = -12

And the given equation expression are J³k/h⁰

Now Substitute the value in the expression then

J³k/h⁰

After putting the values the expression become

(-1)³-12/8⁰

(-1)(-12)/1

After solving the value become

12/1

12.

now the value of expression become is the 12 after put all the values in it.

So, The value of equation which is given is 12.

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Prove that:- (1-sinA)(1+sinA)÷(1+cosA)(1-cosA)=cot^2A

astraxan [27]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 , then

sin²x = 1 - cos²x and cos²x = 1 - sin²x

Consider the left side

$\frac{(1-sinA)(1 + sinA)}{(1+cosA)(1-cosA)}$ ← expand numerator/denominator using FOIL

= $\frac{1-sin^2A}{1-cos^2A}$

= $\frac{1-(1-cos^2A)}{1-(1-sin^2A)}$

= $\frac{1-1+cos^2A}{1-1+sin^2A}$

= $\frac{cos^2A}{sin^2A}$

= cot²A = right side , thus proven

6 0

3 years ago

Convert (-6, -6) from Cartesian coordinates to polar coordinates. Round the length to the nearest 0.01 Express the

Neko [114]

9514 1404 393

Answer:

(8.49; 225°)

Step-by-step explanation:

The angle is a 3rd-quadrant angle. The reference angle will be ...

arctan(-6/-6) = 45°

In the 3rd quadrant, the angle is 45° +180° = 225°.

The magnitude of the vector to the point is its distance from the origin:

√((-6)² +(-6)²) = √(6²·2) = 6√2 ≈ 8.4859 ≈ 8.49

The polar coordinates can be written as (8.49; 225°).

_____

<em>Additional comment</em>

My preferred form for the polar coordinates is 8.49∠225°. Most authors use some sort of notation with parentheses. If parentheses are used, I prefer a semicolon between the coordinate values so they don't get confused with an (x, y) ordered pair that uses a comma. You need to use the coordinate format that is consistent with your curriculum materials.