Answer:
Ram responded, shutting his book and looking at his new friend. ... "Like you see a beautiful flower in a garden, but it is too far from you." She extended ... How does the author support the idea that the dream reflects Anna's reality? 2. The author helps the reader picture how Anna feels by. 3. ... had heard the talking animals.
Step-by-step explanation:
Answer:
D - 24x-16
Step-by-step explanation:
(8×3x) + (8×-2)
The net decrease in calories after walking for 5 hours is 50 calorie.
According to the statement
we have given that the
A 25-year old woman burns 200−20t cal/hr and Her caloric intake from drinking Gatorade is 110t calories during the t hour.
And we have to find the net decrease in calories after walking for 5 hours.
So, For this purpose,
We know that the
calories burns in one hour = 200−20t
calories burns in 5 hours = 5(200−20t)
calories burns in 5 hours = 1000 - 100t
Now,
Calorie intake in t hours = 110t
And
net decrease in calories after walking for 5 hours = 100t - 1000 + 110t
put t is 5 then
net decrease in calories after walking for 5 hours = 210(5) - 1000
net decrease is 50 cal.
So, The net decrease in calories after walking for 5 hours is 50 calorie.
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Answer:
and ![\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%5Capprox%200.661)
Step-by-step explanation:
The equation of the isotope decay is:
![\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }](https://tex.z-dn.net/?f=%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%3D%20e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D)
14-Carbon has a half-life of 5568 years, the time constant of the isotope is:
![\tau = \frac{5568\,years}{\ln 2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B5568%5C%2Cyears%7D%7B%5Cln%202%7D)
![\tau \approx 8032.926\,years](https://tex.z-dn.net/?f=%5Ctau%20%5Capprox%208032.926%5C%2Cyears)
The decay time is:
(There is no a year 0 in chronology).
![t = 3335 \pm 13\,years](https://tex.z-dn.net/?f=t%20%3D%203335%20%5Cpm%2013%5C%2Cyears)
Lastly, the relative amount is estimated by direct substitution:
![\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B%5Cmp%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmin%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmin%7D%20%5Capprox%200.659)
![\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%5Capprox%200.661)
K=9S/2, total savings is S+9S/2=11S/2. So 9S/2÷11S/2=9/11. (a) The ratio is 9:11.
(b) Kieran’s savings are 9/11 of the total savings.
(c) Simon’s savings are S÷11S/2=2/11 of total savings (or simply subtract 9/11 from 1).
(d) K-S=28, 9/11-2/11=7/11 of the total savings, so 7/11 of the total savings is $28 and total savings is 11/7×28=$44.
CHECK
Kieran has 9/11 of $44 which is $36 and Simon has $8. 36/8=9/2 so the ratio is correct.
Simon has $28 less than Kieran and 36-8=28, which is correct.