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egoroff_w [7]
1 year ago
5

What basic trigonometric identity would you use to verify that

Mathematics
1 answer:
Dafna1 [17]1 year ago
6 0

Given the equation:

\frac{\sin^2x+\text{cos}^2x}{\cos x}=\sec x

Let's determine the trigonometric identity that you could be used to verify the exquation.

Let's determine the identity:

Apply the trigonometric identity:

\sin ^2x+\cos ^2x=1

\cos x=\frac{1}{\sec x}

Replace cosx for 1/secx

Thus, we have:

\begin{gathered} \frac{\sin^2x+\cos^2x}{\frac{1}{\sec x}} \\  \\ =(\sin ^2x+\cos ^2x)(\sec x) \\ \text{Where:} \\ (\sin ^2x+\cos ^2x)=1 \\  \\ We\text{ have:} \\ (\sin ^2x+\cos ^2x)(\sec x)=1\sec x=\sec x \end{gathered}

The equation is an identity.

Therefore, the trignonometric identity you would use to verify the equation is:

\cos ^2x+\sin ^2x=1

ANSWER:

\cos ^2x+\sin ^2x=1

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At a family reunion, a family friend brings 20 pounds of ice and 30 cups. Which expression shows the greatest common factor, usi
Eddi Din [679]

The expression for the greatest common factor of 20 and 30 using the distributive property is 10(2 + 3)

Pounds of ice = 20

Number of cups = 30

The relation for the distributive property :

a × (b + c)

Expand ;

a × (b + c) = (a × b) + (a × c) = ab + ac

Finding the greatest common factor of 20 and 30 ;

  • 20 = 2 × 10

  • 30 = 3 × 10

(2 × 10) + (3 × 10)

According to the distributive property

(2 × 10) + (3 × 10) = 10(2 + 3)

Therefore, the expression for the greatest common factor is 10(2 + 3)

Learn more :brainly.com/question/15263211

6 0
3 years ago
4 (x+2)=2 (x+3)+2 (x-1) answer?
vekshin1

4(x+2)=2(x+3)+2(x-1)\\\\use\ the\ distributive\ property:\ a(b\pm c)=ab\pm ac\\\\(4)(x)+(4)(2)=(2)(x)+(2)(3)+(2)(x)-(2)(1)\\\\4x+8=2x+6+2x-2\ \ \ \ |simplify\\\\4x+8=(2x+2x)+(6-2)\\\\4x+8=4x+4\ \ \ \ |-4x\\\\8=4\ FALSE!!!\\\\Answer:\ NO\ SOLUTION\ (x\in\O)

4 0
3 years ago
Please Help and please show the math on how you go the answer
icang [17]

Answer: 1 1/2

Step-by-step explanation:

First subtract 1 1/3 from 3 1/3 to find what is left over from her first batch of browns, 2. Then subtract 1 1/3 from 2 and get 2/3. 2/3 is 1/2 of 1 1/3, so she can make 1 1/2 more batches of brownies.

4 0
3 years ago
Solve for x. Show each step of the solution. 5.5(8-x)+44=104-3.5(3x+24)
earnstyle [38]
5.5 ( 8 - x ) + 44 = 104 - 3.5 ( 3x + 24 )

Distribute 5.5 through the parenthesis.

44 - 5.5x + 44 = 104 - 10.5x - 84

Add the numbers.

88 - 5.5x = 20 - 10.5x

Move variable to the left side and change its sign.

88 - 5.5x + 10.5x = 20
-5.5x + 10.5x = 20 -88

Collect like terms.

5x = 20 - 88

Calculate the sum or difference. 

5x = - 68

Divide both sides by 5.

x = - 68 / 5
4 0
3 years ago
There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the
Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = <u>14.13.12!</u> = <u>14.13 </u>= 91

           12! 2!            2    

C₁₆,₂ = <u>16.15.14!</u> = <u>16.15 </u>= 120

           14! 2!            2    

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = <u>14.13.12.11.10!</u> = <u>14.13.12.11 </u>= 1001

                  10! 4!               4.3.2.1    

C₁₆,₄ = <u>16.15.14.13.12!</u> = <u>16.15.14.13 </u>= 1820

                  12! 4!               4.3.2.1    

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

7 0
3 years ago
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