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MArishka [77]
3 years ago
5

1 : What postulate or theorem can be used to prove △PQR ~ △STU ??

Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

#1 is the AA postulate

Step-by-step explanation:

It is the AA postulate because both triangles have congruent angles, making the triangles similar.

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Can someone help me with this please
MaRussiya [10]
X,y is 5(5y,-2y) and 5(y-2)
3 0
2 years ago
Carrie is finding the set of even numbers within the set of prime numbers. Of the sets described, which is the universal set?
gogolik [260]
Even numbers are considered divisible by 2
Prime numbers are numbers divisible by 1
Considering that in this situation, we only need positive numbers, 2 is the only number. It is even and also prime.

Therefore prime numbers
5 0
2 years ago
I don’t know how to do this. HELP!
Karo-lina-s [1.5K]

Answer:

1/4

Step-by-step explanation:

coins have 2 sides, tails and heads, so for it to land on tails is a 1/2 chance. Dice have 6 sides and 1,3,5 are all odd so its a 3/6 chance. Simplify 3/6 to 1/2 and multiply by 1/2 to get 1/4. im not sure, its my guess.

3 0
3 years ago
A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and
RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

  • A_t = A_0e^r^t
  • where Aₜ = mass at time t, A₀ = mass at time 0,  r = decay constant (rate), t = time  

Our known variables are:

  • 1998 to the year 2004 is a total of t = 6 years.
  • The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

  • 270=360e^-^r^(^6^)
  • 270=360e^-^6^r
  • \frac{3}{4} =e^-^6^r
  • \text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)
  • \text{ln} (\frac{3}{4} )=-6r
  • r=-\frac{\text{ln}\frac{3}{4} }{6}
  • r=0.04794701

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

  • A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)
  • A_t=270e^-^0^.^3^3^5^6^2^9^0^7
  • A_t=193.01982213

The expected mass of the sample in the year 2011 would be 193 mg.

3 0
2 years ago
Help pleaseeeeeeeeeee
timurjin [86]
1 - Right triangle
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2 years ago
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