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
kupik [55]
2 years ago
15

Will mark brainliest and rate 5/5 20 POINTS WILL BE GIVEN

Mathematics
1 answer:
lapo4ka [179]2 years ago
6 0

Answer:

b 5/5 plzzzzzzzzzzzzzzz

You might be interested in
The problem<br> 1/3+1 5/6=
soldi70 [24.7K]
Mixed number form:2 1/6 Decimal form:2.16 exact form: 13/6
4 0
2 years ago
Read 2 more answers
Pls answer……………………..
algol [13]

Answer:

They are not congruent because the lines aren't parallel.

Step-by-step explanation:

For all of those angle relation theorems to be valid, PARALLEL lines need to be split by a transversal.

8 0
2 years ago
Describe the steps you would use to factor 2x3 + 5x2 – 8x – 20 completely. Then state the factored form.
WARRIOR [948]
The polynomial 2x^3 + 5x^2-8x-20 may have solutions which are the divisors of -20, therefore -20 has the following divisors: \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.
If x=1, then 2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0,
if x=-1, then 2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0,
if x=2, then 2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0, then x=2 is a solution and you have the first factor (x-2). 
If x=-2, then 2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0, then x=-2 is a solution, so you have the second factor (x+2).
Since x-2 and x+2 are two factors of 2x^3 + 5x^2-8x-20 , then the polynomial x^2-4 is a divisor of 2x^3 + 5x^2-8x-20 and dividing the polynomial 2x^3 + 5x^2-8x-20 by x^2-4 you obtain 
 2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5).











5 0
3 years ago
Read 2 more answers
I need help on 18 and 19 PLS HELP ME ILY I WILL MARRY U I WILL LOVEEE U FOR ETERNITY
Dmitriy789 [7]

Oof, this is surely a tough one.  But, if I'm looking at this correctly, than I think you were in the right mindset when trying to figure both of them out.  Like, for #18, think about the formula; P(B|A) = P(A and B)/P(A).  So, based on the question, you can let not rolling a 1 on the blue cube be A, and rolling a 1 on the red cube be B.  Now, personally, I prefer using the decimals when heading into the equations, but I'll convert the product into a fraction for the instruction's sake.  0.02777777778 + 0.13888888889 = 1/6.  1/6 ≈ 17.7%.  You can do the same for #19, the overall outcome would be 5/6, which would be about 83%.  Now, that's just me really attempting my hand at it, I will say, Math is far from my strong suite.  But, in all honesty, I really am doing my best to help (and no, it's not because of what you offered, lol, more rather I want to help people who struggle with the same problems I do).  Anyways, I hope this helps, and if you need more information, or anything, just let me know!  Wish ya the best of luck!!  ^u^

5 0
3 years ago
Read 2 more answers
What missing angle of the figure
aniked [119]

Answer:

138

Step-by-step explanation:

152+125=277

277-135=142

around 138

4 0
3 years ago
Other questions:
  • Help help plz about to fail
    11·1 answer
  • Name a pair of supplementary angles. Just any pair of supplementary angles. Plz
    9·1 answer
  • Write the equation (in point slope form) for the line that is perpendicular to the line
    8·1 answer
  • PLEASE HELPPPPPPP!!!!!!!
    6·2 answers
  • Stephanie can paint 4 cars in 1 hour how many cars can she paint in 3.75 hours
    8·1 answer
  • Which of the coordinate points below will fall on a line where the constant of proportionality is 23? Select all that apply.
    7·1 answer
  • Please help it’s due today!!!!!!
    5·1 answer
  • I badly need help with this!!! Pls pls help, thank you
    7·2 answers
  • Pleade god help me find a person that is sweet enough to help me thank you to anyone who will help me
    10·2 answers
  • Describe and correct the error in solving the absolute value inequality.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!