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forsale [732]
3 years ago
11

Need help in these to math problems

Mathematics
1 answer:
Masteriza [31]3 years ago
6 0

Question 1:

For this case we have that by definition, the GCF of two numbers is the largest number that is a factor of both numbers. For example, the number 10 is the biggest factor that 50 and 30 have in common.

We must find the GCF of the following expression:

30t ^ 2u + 12tu ^ 2 + 24tu

We look for the prime factorization of each number:

30: 2 * 3 * 5\\12: 2 * 2 * 3\\24: 2 * 2 * 2 * 3

The GCF of the three numbers is 2 * 3 = 6

So:

6tu (5t + 2u + 4)

Answer:

Option C

Question 2:

For this case we have that by definition of power properties of the same base, that:

a ^ n * a ^ m = a ^ {n + m}

That is, for multiplication of powers of the same base, the same base is placed and the exponents are added.

So:

b ^ n * b ^ m = b ^ {n + m}

ANswer:

b ^ {n + m}

Option A

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How do you find the perimeter and area of this shape?
mixas84 [53]

 

P = 12 + 6 + 8,5 = 18 + 8,5 = 26,5 mm

A = 12 × 4 / 2 = 48 / 2 = 24 mm²



6 0
3 years ago
HELP!!!
Troyanec [42]

Answer:

8

Step-by-step explanation:

18-12

12-X

X=12*12/18

X=8mm

8 0
3 years ago
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
Identify the two prime number that are greater than 25 and less than 35
victus00 [196]
Answer: 29 and 31...
7 0
3 years ago
Read 2 more answers
How many cups of flour do you need for 3 cups of cashews fill in the information in the table. Thanks
Andrei [34K]

Answer:

gh

Step-by-step explanation:

4 0
2 years ago
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