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WITCHER [35]
3 years ago
7

Chord AB is congruent to chord CD. What is the measure of arc CD?

Mathematics
1 answer:
tatiyna3 years ago
8 0
Is there measurements for AB AND CD
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Pls help me. This is a geo quiz and I don’t have a clue what I am doing.
Kruka [31]
3x + 5 + 4x =180
7x + 5 = 180
7x = 175
x = 25


3(25) + 5
75 + 5
80
KL = 80
4 0
3 years ago
PLEASE HELP ME !!!
vfiekz [6]

Answer:

-0.4 or 2.3

Step-by-step explanation:

simple

each box is 0.1 unit

7 0
3 years ago
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ALGUEMMM ME AJUDAAA POR FAVORRR!!!!!!!!!!!!!!!!!!!
Amanda [17]

Answer:

50)

a. + 13

B. +10

C. -15

D. -5

Step-by-step explanation:

8 0
3 years ago
What is the perimeter of the fifth square in this pattern?
torisob [31]
The first square has side lengths of 4 units(because the square root of 16 is 4). The second square has side lengths of 8 units. The third square has side lengths of 12 units. As you can tell, the side lengths increase by 4. So the fourth square will have side lengths of 16 units, and the fifth square will have side lengths of 20 units. The question wants the perimeter, so the answer is 20*4(sum of 4 sides of equal lengths) which is 80 units.
5 0
3 years ago
Billy is monitoring the exponential decay of a radioactive compound. He has a sample of the compound in a test tube in his lab.
Over [174]

Solution:

Formula for radioactive Decay is given by

R_{0}= R(1-\frac{S}{100})^t

R_{0}= Initial Population

R = Remaining population after time in hours

Rate of Decay = S % per hour

Initial Population = 72 grams

Final population = 15 grams

Rate of Decay = 35 % per hour

Substituting the values to get value of t in hours

72=15(1-\frac{35}{100})^t\\\\ 4.8= (0.65)^t\\\\ t= -3.64→→1 St expression

But taking positive value of t , that is after 3.64 hours the sample of 72 grams decays to 15 grams at the rate of 35 % per hour.

Now , it is also given that, Once the sample reaches a mass of 15 grams, Billy will continually add more of the compound to keep the sample size at a minimum of 15 grams.

Substituting these in Decay Formula

Final Sample = 15 gm

Starting Sample = 15 +k, where k is amount of sample added each time to keep the final sample to 15 grams.

Time is over 3.64 hours i.e new time = 3.64 + t

Rate will remain same i.e 35 % per hour.

15=(15+k)(1-\frac{35}{100})^{3.64+t}→→→  Final expression (Second) , that is inequalities can be used to determine the possible mass of the radioactive sample over time.







5 0
3 years ago
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