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lisabon 2012 [21]
3 years ago
7

I need help with pre calculus homework

Mathematics
1 answer:
Vesnalui [34]3 years ago
4 0
Cos^2 + cos^2 tan^2 (break the brackets)

Cos^2+ cos^2 • sin^2/cos^2 (divide out the cos^2)

Cos^2+ sin^2 (Cos2+ sin^2= 1)
= 1
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What is the equation of the line in slope intercept form
Veseljchak [2.6K]

Answer:

y=5/4-10

Step-by-step explanation:

7 0
2 years ago
Write ln(x^3y^2) in terms of u and v (picture provided)
Katena32 [7]

Remember the properties of logarithms. If you have ln81, you can pull out the power of two (9^2) and have 2ln9. Also, you could do ln9 + ln9 because 9*9=81 and ln9*9 = ln9 + ln9

With these properties, you can get 3lnx+2lny, which is 3u+2v according to the answer.

Hope I helped!

6 0
2 years ago
Find the distance between <br><br> (4,3) and (0,3).<br> A) 10 <br> B) 2 <br> C) 2<br> 13<br> D) 4
Daniel [21]

Answer:

D

Step-by-step explanation:

4 units

4 0
3 years ago
STUV~CBED.<br> What is the similarity ratio of STUV to CBED?
ASHA 777 [7]

The similarity ratio of STUV to CBED is 0.5

<h3>How to determine the similarity ratio of STUV to CBED?</h3>

For the shapes to have a similarity ratio, it means that:

The shapes are similar (not necessarily congruent)

From the diagram, the following sides are corresponding sides

ST and CB

Where

ST = 2

CB = 1

The similarity ratio of STUV to CBED is calculated as:

Ratio = CB/ST

Substitute the known values in the above equation

Ratio =1/2

Evaluate

Ratio = 0.5

Hence, the similarity ratio of STUV to CBED is 0.5

Read more about similar shapes at:

brainly.com/question/14285697

#SPJ1

8 0
1 year ago
5.–/1 points SCalcET8 3.8.011. Ask Your Teacher My Notes Question Part Points Submissions Used Scientist can determine the age o
Vilka [71]

Answer:

  • 13,200 years

Explanation:

These steps explain how you estimate the age of the parchment:

1) Carbon - 14 half-life: τ = 5730 years

2) Number of half-lives elapsed: n

3) Age of the parchment = τ×n = 5730×n years = 5730n

4) Exponential decay:

The ratio of the final amount of the radioactive isotope C-14 to the initial amount of the same is one half (1/2) raised to the number of half-lives elapsed (n):

  • A / Ao = (1/2)ⁿ

5) The parchment fragment  had about 74% as much C-14 radioactivity as does plant material on Earth today:

  • ⇒ A / Ao = 74% = 0.74

  • ⇒ A / Ao =  0.74 = (1/2)ⁿ

  • ⇒ ln (0.74) = n ln (1/2)        [apply natural logarithm to both sides]

  • ⇒ n = ln (1/2) / ln (0.74)

  • ⇒ n ≈ - 0.693 / ( - 0.301) = 2.30

Hence, 2.30 half-lives have elapsed and the age of the parchment is:

  • τ×n = 5730n = 5730 (2.30) = 13,179 years

  • Round to the nearest hundred years: 13,200 years
7 0
3 years ago
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