Answer:
5a+5b
Step-by-step explanation:
Multiply each variable in the parenthesis by the number outside.
Hope this helps.
Answer:
Each face of an octahedron is an equilateral triangle. An octahedron is a three-dimensional solid (polyhedron) that has eight faces.
Step-by-step explanation:
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Answer:
the answer is incorrect though the first step is correct but the second step is wrong and so is the answer. the correct answer would be x raise to the power of 2by 5
Step-by-step explanation:
{x⁶/⁵⁻²/⁵} ¹/²= {x⁴/⁵}¹/²= x²/⁵
in the second step the power of the denominator must be subtracted from the power of the numerator using the law of indices then we would get the correct answer
Answer:
450
Step-by-step explanation:
First we have to find the discount
20% of 1500 = 300
if you search up 20 x 1500/100, it would say 300
you can do that same equation in your maths book, and also by canceling out
10% vat tax
so 10% of 1500 = 150
since it is tax, we have to add it to our discount
300 + 150 = 450
450 is the answer
Answer:
Half-life of the goo is 49.5 minutes
![G(t)= 300e^{-0.014t}](https://tex.z-dn.net/?f=G%28t%29%3D%20300e%5E%7B-0.014t%7D)
191.7 grams of goo will remain after 32 minutes
Step-by-step explanation:
Let
denotes initial and final mass.
![M_0=300\,\,grams\,,\,M_f=37.5\,\,grams](https://tex.z-dn.net/?f=M_0%3D300%5C%2C%5C%2Cgrams%5C%2C%2C%5C%2CM_f%3D37.5%5C%2C%5C%2Cgrams)
According to exponential decay,
![\ln \left ( \frac{M_f}{M_0} \right )=-kt](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%20%5Cfrac%7BM_f%7D%7BM_0%7D%20%5Cright%20%29%3D-kt)
Here, t denotes time and k denotes decay constant.
![\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%20%5Cfrac%7BM_f%7D%7BM_0%7D%20%5Cright%20%29%3D-kt%5C%5C%5Cln%20%5Cleft%20%28%20%5Cfrac%7B37.5%7D%7B300%7D%20%5Cright%20%29%3D-k%28150%29%5C%5C-2.079%3D-k%28150%29%5C%5Ck%3D%5Cfrac%7B2.079%7D%7B150%7D%3D0.014)
So, half-life of the goo in minutes is calculated as follows:
![\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%20%5Cfrac%7B50%7D%7B100%7D%20%5Cright%20%29%3D-kt%5C%5C%5Cln%20%5Cleft%20%28%20%5Cfrac%7B50%7D%7B100%7D%20%5Cright%20%29%3D-%280.014%29t%5C%5Ct%3D%5Cfrac%7B-0.693%7D%7B-0.014%7D%3D49.5%5C%2C%5C%2Cminutes)
Half-life of the goo is 49.5 minutes
![\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%20%5Cfrac%7BM_f%7D%7BM_0%7D%20%5Cright%20%29%3D-kt%5CRightarrow%20M_f%3DM_0e%5E%7B-kt%7D)
So,
![G(t)= M_f=M_0e^{-kt}](https://tex.z-dn.net/?f=G%28t%29%3D%20M_f%3DM_0e%5E%7B-kt%7D)
Put ![M_0=300\,\,grams\,,\,k=0.014](https://tex.z-dn.net/?f=M_0%3D300%5C%2C%5C%2Cgrams%5C%2C%2C%5C%2Ck%3D0.014)
![G(t)= 300e^{-0.014t}](https://tex.z-dn.net/?f=G%28t%29%3D%20300e%5E%7B-0.014t%7D)
Put t = 32 minutes
![G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams](https://tex.z-dn.net/?f=G%2832%29%3D%20300e%5E%7B-0.014%2832%29%7D%3D300e%5E%7B-0.448%7D%3D191.7%5C%2C%5C%2Cgrams)