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
damaskus [11]
3 years ago
6

How are these triangles congruent

Mathematics
1 answer:
Sergio039 [100]3 years ago
4 0

Answer:

Step-by-step explanation:

I think it's B

You might be interested in
Describe how you would place 3
zhenek [66]
I would personally take the number three and trow it out the window but that's not what we are talking about
4 0
3 years ago
Grace wants to find the probability of a family of 3 children having 2 boys and 1 girl. Use a tree diagram to list the possibili
pychu [463]

Answer:

Step-by-step explanation:

My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance if you think about it, your answer 2/3 can't be correct. If we assume that boys and girls are born with equal probability, then the probability to have two girls (and one boy) should be the same as the probability to have two boys and one girl. So you would have two cases with probability 2/3, giving an impossible 4/3 probability for both cases. Also, your list "Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy" seems strange. All of those are 2 boys and 1 girl, so based on that list, you should get a 100 percent chance. But what about Boy-Girl-Girl, or Girl-Girl-Girl? You get 2/3 if you assume that adjacencies in the (ordered) list are important, i.e., "2 boys and a girl" means that the girl was not born between the boys.

5 0
2 years ago
The probability that an event will occur is 2/3 what is the best answer of this event likely occurring
Oduvanchick [21]
It's likely that the event will occur.
6 0
3 years ago
(24xy^3-16x^2y^2+32x^2y)/8xy
Ahat [919]
(24xy^3-16x^2y^2+32x^2y)/8xy 

  <span><span>(<span><span>‌<span><span><span><span><span>24x</span><span>y^3</span></span>−<span><span>16<span>x^2</span></span><span>y^2</span></span></span>+<span><span>32<span>x^2</span></span>y</span></span>8</span></span>‌x</span>)</span><span>(y)</span></span><span>  =<span><span><span>−<span><span>2<span>x^3</span></span><span>y^3</span></span></span>+<span><span>3<span>x^2</span></span><span>y^4</span></span></span>+<span><span>4<span>x^3</span></span><span>y^<span>2</span></span></span></span></span>

4 0
3 years ago
A box can had a maximum of 60 comic books.If comic books are bundled together in groups of 8, write and solve an inequality to f
Reika [66]

Answer: the maximum number of bundles of comic books that the box can carry is 7

Step-by-step explanation:

Let x represent the maximum number of bundles of comic books that the box can carry.

If comic books are bundled together in groups of 8, then the total number of comic books in x bundles of comic books would be 8x.

A box can hold a maximum of 60 comic books. Therefore, the inequality to find the maximum number of bundles of comic books that the box can carry is

8x ≤ 60

x ≤ 60/8

x ≤ 7.5

8 0
3 years ago
Other questions:
  • A salesperson earns $200 a week plus a 4% commission on her sales. Which equation models the relationship between her sales in a
    9·2 answers
  • Put these fractions in order from smallest to greatest<br><br> 3/4, 2/7, 2/10, 5/8, 2/3, 8/4
    10·1 answer
  • Am I right so far?? Idk pls help
    9·1 answer
  • The difference between four times a number and five has a result of negative eight.
    10·2 answers
  • Pls help need the answers quick ?!?!?!
    15·2 answers
  • 2 Simplify -3(16 = 35+4)/5. Show your work.
    10·1 answer
  • PLEASE HELP ME!!! Find m (angle) LMN
    8·1 answer
  • I need help! anyone know this?
    12·1 answer
  • You find
    5·1 answer
  • Helppp plzzzzzzz..………..
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!