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maxonik [38]
2 years ago
13

The first one will get the brianlist please help answer these two

Mathematics
2 answers:
Julli [10]2 years ago
8 0

Answer:

<em>Hello bakugo here!</em>

Step-by-step explanation:

I don't see anything, is there supposed to be anything here? if not

<em>Thank you extra for FREE points! :D</em>

viktelen [127]2 years ago
5 0
Yeah I also don’t see anything
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K, you have 2 minutes, one minute for one kilometer.
If you're in 60km/h, you're on 1km/m, so, if you cover 1 km in one minute, you need to stay in that speed to get to the train station. 
4 0
3 years ago
A large mixing tank initially contains 1000 gallons of water in which 40 pounds of salt have been dissolved. Another brine solut
sergey [27]

Answer:

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

Step-by-step explanation:

Given the data in the question;

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?

dA/dt = rate in - rate out

first we determine the rate in and rate out;

rate in = 3pound/gallon × 5gallons/min = 15 pound/min

rate out = A pounds/1000gallons × 5gallons/min  = 5Ag/1000pounds/min

= 0.005A pounds/min

so we substitute

dA/dt = rate in - rate out

dA/dt =  15 - 0.005A

Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

8 0
3 years ago
Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
Can someone please help on this problem and how it is done? Picture is attached. Thank you​
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The angle passes 90 degrees plus an additional 55 degrees
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