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2 years ago

Helloppp i need hwlp with this plllss

Mathematics

1 answer:

Rasek [7]2 years ago

3 0

the answer is b because x is greater than or equal to 0

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

If the height is 12 inches and the diameter is 16 inches, what is the volume?

tigry1 [53]

Answer:

V = πr²h

V = π(8)^2×12 r = ½diameter= 16/2=8

V = 64π×12

V= 64 × 3.14 ×12

V = 2,411.52 cubic inches.

Hope it helps!

4 0

2 years ago

1/8 (3d-2) = 1/4 (d+5)

tatiyna

OK, so for this equation, your goal is to get the d, and ONLY the d, on one side of the equation. So, to start out, you need to multiply the entire equation, meaning both sides, by 8 because we are trying to get rid of those pesky fractions.
8(1/8(3d-2)=1/4(d+5))
The equation then turns into this because the 8 and 4 cancelled out with the 8.
1(3d-2)=2(d+5)
Now, we need to distribute the left over numbers into the parenthesis.
3d-2=2d+10
And finally, we need to get the d's on one side, and the numbers on the other, so we subtract 2d from both sides and add the 2 to both sides. They then cancel out to make
d=12
Hope it helps! :)

8 0

3 years ago

In the equation y = mx + b, the "b" is the y-intercept<br><br> True<br> False

DENIUS [597]

Answer:

False

Hope my answer helps you pls mark me as brainiest :)

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3 years ago

When writing a linear equation in standard form, what other forms might you need to use?

wlad13 [49]

Answer: