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
frutty [35]
3 years ago
15

22 + (-41) How you do dis

Mathematics
1 answer:
Irina18 [472]3 years ago
6 0
-19
Explanation: 22-41=-19
You might be interested in
Write the expression in simplest radical form.<br><br> √75<br><br> √75<br> 3√5<br> 5√3<br> 15
Wittaler [7]
Go to Mathway .com they'll help you out and do it for you 
that way you dont have to wait
4 0
4 years ago
75 = 17 ÷2×93871788=?<br>?÷748267818=?<br>?+7376818938÷7477818199×89=?​
Alchen [17]

Answer:

YOUR MOM

Step-by-step explanation:

LOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLO)LOLOLOLOLLOllll

6 0
3 years ago
Find the measure of a dilated image of segment CD, 5 units long, with a scale factor of 4?
Amiraneli [1.4K]
The measure would be 20 since five times four is 20
8 0
3 years ago
How do i find the area of a trapezoid? please make your answer simple yet descriptive...
Vera_Pavlovna [14]
A = a+b/2 x h
Add sides a and b then divide by 2
After that multiply by the h
4 0
4 years ago
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
Other questions:
  • A young equestrian loved to ride her horse all day long, dreaming of the day she would ride in the Kentucky Derby. One day in ea
    8·1 answer
  • Which fraction is equivalent to negative 5 over 9?
    9·2 answers
  • Before tax, and miscellaneous charges, Jason's cell phone bill is $100 per month plus $0.15 for every text message that he sends
    11·2 answers
  • Caleb purchases 5 beach towels for d dollars each. Each towel is charged an additional 6% sales tax.
    7·1 answer
  • My tens digit is double my ones digit. I am less than 70 and greater than 60.
    13·1 answer
  • Place the following numbers in order from least to greatest.<br>3.2 , 115%, 0.09, 5/12, 2/7​
    13·1 answer
  • 2 Points<br> What can you say about the end behavior of the function f(x)=-4x + 6x2 - 52?
    6·1 answer
  • Solve the inequality −24z≥−13
    8·2 answers
  • A marble is selected at random from a jar containing 4 red marbles, 5 yellow marbles, and 3 green marbles. What is the probabili
    13·1 answer
  • The square of a certain number plus ten is equal to 81​
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!