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Y_Kistochka [10]
2 years ago
5

VERY EASY, WILL GIVE 50 POINTS FOR CORRECT ANSWER ASAP AND WILL GIVE BRAINLIEST.

Mathematics
1 answer:
Murrr4er [49]2 years ago
5 0

Answer:

D) R’(2,3); S’(7,3); T’(5,8); U’(0,8)

Step-by-step explanation:

The last one.  If you reflect it over the x-axis, it moves from quadrant IV to quadrant I.  Everything in quadrant I is positive, and the last one is the only one that has all x and y values positive.  

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Find the circumference of the circle. <br> A. 11 cm<br> B. 44 cm<br> C. 22 cm<br> D. 33 cm
ira [324]

Answer:

The answer is 44 cm .

which is (B)

5 0
3 years ago
How would negative 465 be classified
DENIUS [597]
As an integer and a raitonal number as well as a real number.
6 0
3 years ago
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Suppose that you and a friend are playing cards and you decide to make a friendly wager. The bet is that you will draw two cards
Dmitrij [34]

Answer:

Ok, the probability of winning is:

For the first draw, we have 13 diamonds in a 52 card deck, the probability of drawing a diamond is:

p1 = 13/52

for the second draw we have 12 diamonds and 51 cards in the deck, now the probability is:

p2 = 12/51

the total probability is equal to the product of the individual probabilities, this is:

P = p1*p2= (13/52)*(12/51) = 0.059

now, the expected value can be calculated as follows.

EV = (P1*X1 + P2*X2)

where P1 is the probabilty of event 1, and X1 is the event 1,

In this case X1 is the event of winning $29, then P1 = 0.059

X2 will be the event of loosing $4, so P2 = 1 - 0.059 = 0.941

Then the expected value for 1 round of the game is:

EV = (0.059*$29 - 0.941*$4) = -$2.05

So if you play 891 times, expected value of 891 times the expected value for 1 round, this is:

891*EV = 891*(-$2.05) = -$1826.55

So you loss around  $1826.55

4 0
3 years ago
11,647 rounded to the nearest hundred
fenix001 [56]

Answer:

11,600

Step-by-step exp

the 4 in the tens place means the 6 stays the same

6 0
3 years ago
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I’m lost please help
vagabundo [1.1K]

Answer:

See proof below

Step-by-step explanation:

Two triangles are said to be congruent if one of the 4 following rules is valid

  1. The three sides are equal
  2. The three angles are equal
  3. Two angles are the same and a corresponding side is the same
  4. Two sides are equal and the angle between the two sides is equal

Let's consider the two triangles ΔABC and ΔAED.

ΔABC sides are AB, BC and AC

ΔAED sides are AD, AE and ED

We have AE = AC and EB = CD

So AE + EB = AC + CD

But AE + EB = AB and AC+CD = AD

We have

AB of ΔABC  = AD of ΔAED

AC of ΔABC =  AE of ΔAED

Thus two sides the these two triangles. In order to prove that the triangles are congruent by rule 4, we have to prove that the angle between the sides is also equal. We see that the common angle is ∡BAC = ∡EAC

So  triangles ΔABC and ΔAED are congruent

That means all 3 sides of these triangles are equal as well as all the angles

Since BC is the third side of ΔABC and ED the third side of ΔAED, it follows that

BC = ED  Proved

5 0
1 year ago
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