1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Damm [24]
3 years ago
5

(44 points and brainliest) A geologist noticed that land along a fault line moved 24 4/5 centimeters over the past 175 years. On

average, how much did the land move each year?
Mathematics
2 answers:
miskamm [114]3 years ago
5 0
24.8 cm ÷ 175= 0.141... or 0.14 cm.
The answer is 0.14 cm.

Hope this helps
lakkis [162]3 years ago
3 0

Answer:

7.0564516129

Step-by-step explanation:

Use 175 and divide 24 4/5 from that, then you find out how much the land moves each year.

You might be interested in
If sin theta = 8/17 and cot theta < 0, what is sec theta?
klasskru [66]

Answer:

-\frac{17}{15}

Step-by-step explanation:

By definition, \cot \theta=\frac{1}{\tan \theta} and \sec \theta=\frac{1}{\cos \theta}. Since since \cot \theta is negative, \tan \theta must also be negative, and since \sin \theta is positive, we must be in Quadrant II.

In a right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. The cosine of an angle in a right triangle is equal to its adjacent side divided by the hypotenuse. Therefore, we can draw a right triangle in Quadrant II, where the opposite side to angle theta is 8 and the hypotenuse of the triangle is 17.

To find the remaining leg, use to the Pythagorean Theorem, where a^2+b^2=c^2, where c is the hypotenuse, or longest side, of the right triangle and a and b are the two legs of the right triangle.

Solving, we get:

8^2+b^2=17^2,\\b^2=17^2-8^2,\\b^2=\sqrt{17^2-8^2}=\sqrt{225}=15

Since all values of cosine theta are negative in Quadrant II, all values of secant theta must also be negative in Quadrant II.

Thus, we have:

\sec\theta=\frac{1}{\cos \theta}=-\frac{1}{\frac{15}{17}}=\boxed{-\frac{17}{15}}

6 0
3 years ago
I showed you a curvilinear relationship between age and memory. Now, I want you come up with two variables that you think might
Bogdan [553]

Answer:

Step-by-step explanation:

Curvilinear relationship

A curvilinear relationship is a type of relationship in which there are two variables. As and when the value of one variable increases, so does the value of the other. This continues until a certain point, after which an increase in one variable decreases the value of the other.

Example:

The two variables are - Work pressure and work performance. As work pressure increases, work performance increases until a certain point. After a threshold, when work pressure exceeds, work performance drops. This results in a curvilinear relationship between the two variables.

4 0
3 years ago
B. After a 25% discount, the price of a T-shirt was $12. What was the price before
Kipish [7]

The price before  the discount is $ 16

The population last year is 5280

<h3><u>Solution:</u></h3>

<u><em>B) After a 25% discount, the price of a T-shirt was $12. What was the price before  the discount?</em></u>

Given that after a 25% discount, the price of a T-shirt was $12.  

We have to find what was the price before the discount?  

Now, we know that,

original price – discount = selling price

Original price – 25% of original price = $12

Let the original price be "a"

\begin{array}{l}{\mathrm{a}-25 \% \text { of } \mathrm{a}=12} \\\\ {a-\frac{25}{100} \times a=12} \\\\ {a\left(1-\frac{25}{100}\right)=12} \\\\ {a\left(1-\frac{1}{4}\right)=12} \\\\ {a\left(1-\frac{1}{4}\right)=12} \\\\ {a\left(\frac{3}{4}\right)=12} \\\\ {a=12 \times \frac{4}{3}=16}\end{array}

Thus the original price is $ 16

<em><u>C) Compared to last year, the population of Boom Town has increased by 25%. The  population is now 6,600. What was the population last year?</u></em>

Last year population + 25% of last year population = present population.

\begin{array}{l}{ \text { Last year population( }100 \%+25 \%)=6600} \\\\ {\text { Last year population } \times 125 \%=6600} \\\\ {\text { Last year population } \frac{125}{100}=6600} \\\\ {\text { Last year population }=6600 \times \frac{4}{5}} \\\\ {\text { Last year population }=660 \times 2 \times 4} \\\\ {\text { Last year population }=5280}\end{array}

Thus the last year population is 5280

4 0
4 years ago
Which of the following figures is not a polygon? <br><br> A. <br> B. <br> C. <br> D.
Alexus [3.1K]
B. polygons are figures with 3 sides or more
8 0
3 years ago
Read 2 more answers
55/5 divided by 11/4
Anuta_ua [19.1K]

Answer:

1/4 in fraction form or in decimal form its 0.25

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
Other questions:
  • The area of Jackson's rectangular garden is 64 square ft. If the length of the garden is 16 ft, what is the width of the garden?
    12·2 answers
  • Any help with this:) hope I get help
    6·1 answer
  • Nina paid $450 for 9 shirts. At this rate, how much would 12 shirts cost?
    13·2 answers
  • 8. Maria practiced for her piano recitaleach day for three days. The first day shepracticed for hour, the second day shepractice
    6·1 answer
  • Hi people, Please if someone can give me a hand, l already have done the first part of the exercise, but l cant make Angle CAB=
    6·1 answer
  • (Please show all work)
    8·1 answer
  • Help please !!! I never understood this
    5·1 answer
  • List a typical value for each distribution. Explain how you found the answer.
    6·1 answer
  • What is the slope of the line shown below?
    7·2 answers
  • Question 1 of 10
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!