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Lorico [155]
2 years ago
12

Solve the equation by first using a sum-to-product formula. (enter your answers as a comma-separated list. let k be any integer.

round terms to three
Mathematics
1 answer:
kykrilka [37]2 years ago
8 0

Solutions of the equation are 22.5°, 30°.

The given equation is sin(5θ) - sin(3θ) = cos(4θ)

We take left side of the equation

sin(5θ) - sin(3θ) = 2cos ((5θ+3θ)/2) (sin(5θ-3θ)/2)

=2cos4θsinθ  [From sum-product identity]

Now we can write the equation as

2cos(4θ)sin(θ) = cos(4θ)

2cos(4θ)sinθ - cos(4θ) = 0

cos(4θ)[2sinθ - 1] = 0

cos(4θ) = 0

4θ = 90°

θ = 90/4

θ = 22.5°

and (2sinθ - 1) = 0

sinθ = 1/2

θ = 30°

Therefore, solutions of the equation are 22.5°, 30°

For more information about trigonometric identities, visit

brainly.com/question/24496175

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April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3
vlada-n [284]

Answer:

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = 5\frac{1}{4}\ hrs

5\frac{1}{4}\ hrs can be Rewritten as \frac{21}{4}\ hrs

Number of Hours Carl worked on Math project = \frac{21}{4}\ hrs

Number of Hours Sonia worked on Math project = 6\frac{1}{2}\ hrs

6\frac{1}{2}\ hrs can be rewritten as \frac{13}{2}\ hrs

Number of Hours Sonia worked on Math project = \frac{13}{2}\ hrs

Number of Hours Tony worked on Math project = 5\frac{2}{3}\ hrs

5\frac{2}{3}\ hrs can be rewritten as \frac{17}{3}\ hrs.

Number of Hours Tony worked on Math project = \frac{17}{3}\ hrs.

Now Given:

April worked 1\frac{1}{2} times as long on her math project as did Carl.

1\frac{1}{2}  can be Rewritten as \frac{3}{2}

Number of Hours April worked on math project = \frac{3}{2} \times Number of Hours Carl worked on Math project

Number of Hours April worked on math project = \frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs

Also Given:

Debbie worked 1\frac{1}{4} times as long as Sonia.

1\frac{1}{4}  can be Rewritten as \frac{5}{4}.

Number of Hours Debbie worked on math project = \frac{5}{4} \times Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = \frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs

Also Given:

Richard worked 1\frac{3}{8} times as long as tony.

1\frac{3}{8} can be Rewritten as \frac{11}{8}

Number of Hours Richard worked on math project = \frac{11}{8} \times Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = \frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs

Hence We will match each student with number of hours she worked.

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

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ANSWER: r=18

EXPLANATION:

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What is the answer to 11=7x+2-4x
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Answer:

x=3

Step-by-step explanation:

11=7x+2-4x

=7x+2-4x=11

=3x+2=11

=3x=9

x=3

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3 years ago
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If my current grade is a 65.50 percent and I get an 98 percent on a 60 point test what will be my new percentage
ioda

Answer:

81.75%

Step-by-step explanation:

98 percent on a 60 point test is 59/60

65.50+98/2 = 81.75%

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Example 2 Determine the sum of the first 100 terms for the series 2+5+ 8 + ...​
balandron [24]

Answer:

15050

Step-by-step explanation:

Hello!

So basically, this arithmetic sequence follows the rule (3n-1). What you are looking for is the summation of (3n-1) with a lower limit of 1 and an upper limit of 100. This, plugging it into a calculator (etc. symbolab) or even calculating it manually, we get ∑↑100↓n=1⇒3n-1=15050.

5 0
3 years ago
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