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Viefleur [7K]
4 years ago
15

More comparing decimals

Mathematics
1 answer:
Lisa [10]4 years ago
5 0

When comparing, think of the signs as a crocodile. A crocodile will want to eat greater (bigger) numbers than smaller numbers. So, you can think of it as that way.

2. 0.5 < 0.8 ... 0.8 is greater than 0.5

5. 0.7 = 0.70 ... This is same. 0.70 just has an extra 0 which you don't need.

8. 2.9 > 2.8 ... 2.9 is greater since 9 is bigger than 8.

11. 0.65 > 0.36 ... 65 is greater than 36. Therefore, 0.65 is greater.

14. 4.50 > 4.05 ... 4.50 has a five in the tenth place. 4.05 has five in the hundredth place.

17. 6.01 < 6.1 ... 6.1 has a one in the tenth place. 6.01 has a one in the hundredth place.

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vazorg [7]

Answer:

203.75

Step-by-step explanation:

815 ÷ 4 = 203.75

3 0
3 years ago
Select the correct answer.
Finger [1]

Answer:

Step-by-step explanation: 4a³ + 24ab² - 25b³

(In other words B)

4 0
1 year ago
⚠NEED HELP ASAP! IM COUNTING ON YOU GUYS!⚠(⊙_⊙;) 30 POINTS! WILL MARK BRAINLIEST IF YOUR ANSWER IS CORRECT!
Serhud [2]

Answer:

1) 8%

2) 18 groups

3) b. 23 groups

Step-by-step explanation:

2/25 × 100 = 8%

One each group: 2/25

2/25 × 224 = 17.92

Atleast 2 freshmen: 5

5/25 × 115 = 23

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Artyom0805 [142]

Answer:

what grade level is this

5 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
3 years ago
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