Ask question

Ask question

All categories

2 years ago

12

Answer the 5 questions Listed below Please Show your Work

1 answer:

Zolol [24]2 years ago

5 0

(6 1/3) / (5/6) = (19/3) / (5/6) = 19/3 * 6/5 = 114/15 = 38/5

3 - 7^2 + 3 * 4^2
3 - 49 + 3 * 16
3 - 49 + 48
- 46 + 48
2

(7b - 2) / (-a + 1)....a = -2, b = 3, c = -1/3
where is he c in this equation ? Because if I just use a and b, my answer is not an answer choice

(7b - 10) / (a - 1)...a = -1, b = 5, c = -2/3
same as before...where is the c in this equation

V = (pi) r^2* h
V = (pi)(5^2)(7)
V = (pi)(25)(7)
V = 175(pi)

You might be interested in

Determine the zeros of the function f(x)=x^2-2x-8.

ozzi

X = 4, -2 that should be the roots(zeros) of the function.

6 0

2 years ago

Two fifths of the senior class earned a grade of B+ or higher on an advanced mathematics exam. If 42 seniors scored at least a B

g100num [7]

Answer:

Step-by-step explanation:

its is 105

5 0

2 years ago

Read 2 more answers

What is the value of the expression 8 cubed?

Ierofanga [76]

Answer:

512

Step-by-step explanation:

8*8*8

64*8

512

5 0

2 years ago

Read 2 more answers

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

Find the values for a, b, and c that complete the simplification.

Ira Lisetskai [31]

Answer:

a = 6, b = 4, c = 2

Step-by-step explanation

see attached

3 0

3 years ago

Read 2 more answers