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
Fiesta28 [93]
3 years ago
10

Which transformations could have occurred to map ABC

Mathematics
1 answer:
antoniya [11.8K]3 years ago
5 0
A translation and dilation
You might be interested in
What are the x- and y-intercepts for the graph of 5x – 2y = 20?
Elza [17]

Answer:

x-intercept = 4

y-intercept = −10

Step-by-step explanation:

x-intercept: Plug 0 in for yto get this:

5x = 20 \\ \\ 4 = x

y-intercept: Plug 0 in for xto get this:

-2y = 20 \\ \\ -10 = y

I am joyous to assist you anytime.

3 0
3 years ago
Read 2 more answers
Need help with geometry question?
Nuetrik [128]

Answer: A, Corresponding; Vertical


Step-by-step explanation: The answer for angles 1 and 5 are Corresponding Angles (Elevator Angles)  and angles 1 and 4 are Vertical Angles.

All Angles besides a Linear Pair and Same-Side Interior Angle are Congruent.

5 0
3 years ago
A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang
laiz [17]

<u>Answer-</u>

a. Probability that  three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

\binom{19}{3} = \frac{19!}{3!16!} = 969

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

\binom{33}{6} = \frac{33!}{6!27!} =1107568

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

\binom{52}{9} = \frac{52!}{9!43!} = 3679075400

∴P(3 white candies) = \frac{1073233392}{3679075400} =0.29



Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}

=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})

=(969)(45)(7)(5)(15)=22892625

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

\binom{52}{9}=\frac{52!}{9!43!} =3679075400

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

\frac{22892625}{3679075400} = 0.006


6 0
3 years ago
Can someone plz help
erica [24]

Answer:

4th choice which is 50.24 sq.m

4 0
2 years ago
Read 2 more answers
If David plays 3 tennis matches every week for 9 weeks, how many matches will he play altogether
kari74 [83]
If he plays 3 matches a week for 9 weeks, you have to multiply 3&9.
3*9=27
Daniel will have played 27 tennis matches by the end of the 9 weeks.
Hope this helps:)
5 0
3 years ago
Read 2 more answers
Other questions:
  • Laura wants to show 70 in tens. How many tens will she draw? How do you know?
    8·2 answers
  • Find the amplitude y = 5 sin one half x.
    5·1 answer
  • Please help it would be appreciated!
    5·1 answer
  • A train travels 304 miles in 4 hours if it continues to travel at the same rate,how many miles willit travel 7.5 hours?
    8·1 answer
  • A rectangular field that measures 32 feet by 14 feet must be increased by a
    11·1 answer
  • What is the angle in both radians and degrees determined by an arc of length 2pi meters on a circle of radius 8 meters?
    12·1 answer
  • 2.
    15·1 answer
  • Need help on this.what's the value of a b and c??<br>​
    6·1 answer
  • Domain and range pls
    11·2 answers
  • S = 4w + 12 Which is the CORRECT graph for this equation?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!