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
Elina [12.6K]
3 years ago
5

A^8/a^3 hey can someone help me

Mathematics
1 answer:
Stells [14]3 years ago
3 0
A^8/a^3=a^5 you subtract the exponents
You might be interested in
A club has 12 members. In how many ways can we select four members to go on a trip?
ivanzaharov [21]
This is the number of combinations of 4 from 12

12C4   =        12!
                 ----------
                  4! 8!

=    12*11*10*9             11*5*9  =  495
      --------------     = 
        4*3*2*1
3 0
3 years ago
The length of a rectangle is 4 m less than the diagonal and the width is 5 m less than the diagonal. If the area is 82 m^2, how
nikklg [1K]

I hate rounding.

Let's call the diagonal x.  It's the hypotenuse of the right triangle whose legs are the rectangle sides.

According to the problem we have a length x-4 and a width x-5 and an area

82 = (x-4)(x-5)

82 = x^2 - 9x + 20

0 = x^2 - 9x - 62

That one doesn't seem to factor so we go to the quadratic formula

x = \frac 1 2(9 \pm \sqrt{9^2-4(62)}) = \frac 1 2(9 \pm \sqrt{329})

Only the positive value makes any sense for this problem, so we conclude

x = \frac 1 2(9 \pm \sqrt{329})

That's the exact answer.  Did I mention I hate rounding?  That's about

x = 13.6 meters

Answer: 13.6

----------

It's not clear to me this problem is consistent.  By the Pythagorean Theorem the diagonal satisfies

x^2 = (x-4)^2 + (x-5)^2

which works out to

x=9 \pm 2\sqrt{10}

That's not consistent with the first answer; this problem really has no solution.  Tell your teacher to get better material.

5 0
3 years ago
46 x 53 please show with work or something
lutik1710 [3]

Answer:

2,438

Step-by-step explanation:

    53

    46

______

     318

   212

_______

   2438

6 times 3 is 18. Put the 8 in the ones place and carry the 1 ~ 6 times 5 is 30. Add the carried 1 and you have 31. Put that next to the 8. ~ 4 times 3 is 12. put the 2 under the 1 in 318 and carry the 1. ~ 4 times 5 is 20. Add the carried 1 and you have 21. ~ Put that next to the 2. Now add 318 to 2120 and you have the answer of 12438.

5 0
3 years ago
What is 1997 from 2014????????????
katen-ka-za [31]
What? That doesn't make sence! Any way if you mean 1997 to 2014 how many years.
3 0
3 years ago
Read 2 more answers
5. Xian plans to run 14 laps around a track Each lap is 400 yards. So farxian has run 1,680 yards around the track. What percent
Igoryamba
1,680 divided by 400 = 4.2 
So we know that Xian has done 4.2 laps. 
4.2 divided by 14 = 0.3
0.3 x 100 = 30%
Therefore, Xian has finished 30% of the run
6 0
3 years ago
Other questions:
  • 0.12 or 1.2% which is greater
    8·2 answers
  • Alexia made a down payment of $1825 on a car, and then paid 36 monthly payments of $415. If the loan had 0% interest, how much d
    6·1 answer
  • On any given day, Buffy is either cheerful (c), so-so (s), or gloomy (g). If she is cheerful today, then she will be c, s, or g
    8·1 answer
  • in febuary you have a balance of $270 in your bank account. each month you deposit $45. let january=1, febuary=2 and so on. brai
    14·1 answer
  • What does point A represent in this box plot?
    10·1 answer
  • Jason has a coupon for $2.50 off an electronic book from an online bookstore. If the store original price in dollars of an elect
    10·2 answers
  • Set up an appropriate equation and solve. Data are accurate to two significant digits.
    7·1 answer
  • What is the least common denominator for the fractions
    9·1 answer
  • A new roller coaster is set to open at an amusement park. Due to the height and speed of the roller
    8·1 answer
  • Katherine is younger than Alexandra. Their ages are consecutive integers.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!