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
SVEN [57.7K]
3 years ago
13

What is the measure of each angle in the circle? 60°. 90° 72° 30°

Mathematics
2 answers:
Triss [41]3 years ago
9 0

Hope this helps you

1)

Answer: 45

360 divided by 8 = 45

2) 72

360 divided by 5 = 72

Lady bird [3.3K]3 years ago
4 0
60 degrees hope it helped
You might be interested in
You have a credit card with a balance of $754.43 at a 13.6% APR. You have $300.00 available each month to save or pay down your
pochemuha
 <span>B(n) = A(1 + i)^n - (P/i)[(1 + i)^n - 1] 

where B is the balance after n payments are made, i is the monthly interest rate, P is the monthly payment and A is the initial amount of loan. 

We require B(n) = 0...i.e. balance of 0 after n months. 

so, 0 = A(1 + i)^n - (P/i)[(1 + i)^n - 1] 

Then, with some algebraic juggling we get: 

n = -[log(1 - (Ai/P)]/log(1 + i) 

Now, payment is at the beginning of the month, so A = $754.43 - $150 => $604.43 

Also, i = (13.6/100)/12 => 0.136/12 per month 

i.e. n = -[log(1 - (604.43)(0.136/12)/150)]/log(1 + 0.136/12) 

so, n = 4.15 months...i.e. 4 payments + remainder 

b) Now we have A = $754.43 - $300 = $454.43 so, 

n = -[log(1 - (454.43)(0.136/12)/300)]/log(1 + 0.136/12) 

so, n = 1.54 months...i.e. 1 payment + remainder 
</span>
8 0
4 years ago
Ana tiene 14 años menos que Beto y ambas edades suman 56 años
allsm [11]
42 because 56 mines 14 equals 42
7 0
3 years ago
7 + 14 greatest common factor
boyakko [2]

Answer:

The greatest common factor is 21,

Step-by-step explanation:

Find the prime factors of each term in order to find the greatest common factor (GCF).

8 0
3 years ago
Read 2 more answers
seven different gifts are to be distributed among 10 children. How many distinct results are possible if no child is to receive
Alik [6]

Answer:

120 distinct results are possible if no child is to receive more than one gifts.

Step-by-step explanation:

When the order is not important, we use the combination formula:

C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

The order is said to be not important if for example, John receiving the Buffalo Bills jersey and then Laura receiving the Cleveland Browns jersey is the same as Laura receiving the Cleveland Browns jersey before John receives the Buffalo Bills jersey.

In this problem, we have that:

Combinations of 7 from a set of 10 elements. So

C_{n,x} = \frac{n!}{x!(n-x)!}

C_{10,7} = \frac{10!}{7!(3)!} = 120

120 distinct results are possible if no child is to receive more than one gifts.

6 0
3 years ago
A boat traveled 60 miles downstream and back. The trip downstream took 3 hours. The trip back
Katen [24]
The trip back the boat will travel only 30 miles because the trip doubled the hrs.
5 0
3 years ago
Read 2 more answers
Other questions:
  • Write each number as a product using the GCF as a​ factor, and apply the Distributive Property.
    14·1 answer
  • Select the correct answer.
    11·1 answer
  • Read a frequency graph help please!
    12·1 answer
  • 6у — 10+ 6у + 10=180​
    12·2 answers
  • Someone please help me on this question
    13·1 answer
  • A group of tourists go on a tour. the tour guide rents 15 vans. each van holds 9 tourists write a division problem that can be u
    11·1 answer
  • Which table represents a nonlinear function? pls hurry
    8·1 answer
  • Kyla runs around a 400-meter track at a constant speed of 200 meters per minute. How many
    11·2 answers
  • What is the volume of the rectangular pyramid shown​
    8·1 answer
  • A company is to add handmade bottles and wallets to its product line. Each bottle
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!