We are given with the rate of change of the base, db/dt equal to 1 ft/ sec and the rate of change of the height of the triangle, dh/dt equal to 2 ft/sec. b is 10 ft and h is 70 ft. Area of triangle is equal to A= 0.5 bh The rate of change of the area is equal to dA/dt = 0.5 b dh/dt + 0.5 h db/dt. Substituting, dA= 0.5*10*2 + 0.5*70 * 1 equal to 45 ft2 / sec.
Given:
Consider the equation is:

To prove:
by using the properties of logarithms.
Solution:
We have,

Taking left hand side (LHS), we get

![\left[\because \log_ab=\dfrac{\log_x a}{\log_x b}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_ab%3D%5Cdfrac%7B%5Clog_x%20a%7D%7B%5Clog_x%20b%7D%5Cright%5D)

![[\because \log x^n=n\log x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog%20x%5En%3Dn%5Clog%20x%5D)

![\left[\because \log_ab=\dfrac{\log_x a}{\log_x b}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_ab%3D%5Cdfrac%7B%5Clog_x%20a%7D%7B%5Clog_x%20b%7D%5Cright%5D)

Hence proved.
Answer:
Step-by-step explanation:So we know x + y = 21 and 5x +10y = 165 We line them up in columns x + y = 21 5x +10y = 165 To eliminate the x variable, I'll multiply every element in the 1st equation by -5. -5x + -5y = -105 5x + 10y = 165 Now we combine (add) the equations, which eliminates x altogether. 5y = 60 ... y = 12 From there, x + 12 = 21 ... x = 21-12 ... x = 9 You should double check. Do 9 nickles and 12 dimes equal $1.65? A very important thing to remember using this method is to do the same thing to each element in the equation that you change! I hope that helps.
no se pero de seguro alguien más te ayudará
Answer:
a)0.7
b) 10.03
c) 0.0801
Step-by-step explanation:
Rate of return Probability
9.5 0.1
9.8 0.2
10 0.3
10.2 0.3
10.6 0.1
a.
P(Rate of return is at least 10%)=P(R=10)+P(R=10.2)+P(R=10.6)
P(Rate of return is at least 10%)=0.3+0.3+0.1
P(Rate of return is at least 10%)=0.7
b)
Expected rate of return=E(x)=sum(x*p(x))
Rate of return(x) Probability(p(x)) x*p(x)
9.5 0.1 0.95
9.8 0.2 1.96
10 0.3 3
10.2 0.3 3.06
10.6 0.1 1.06
Expected rate of return=E(x)=sum(x*p(x))
Expected rate of return=0.95+1.96+3+3.06+1.06=10.03
c)
variance of the rate of return=V(x)=![sum(x^2p(x))-[sum(x*p(x))]^2](https://tex.z-dn.net/?f=sum%28x%5E2p%28x%29%29-%5Bsum%28x%2Ap%28x%29%29%5D%5E2)
Rate of return(x) Probability(p(x)) x*p(x) x²*p(x)
9.5 0.1 0.95 9.025
9.8 0.2 1.96 19.208
10 0.3 3 30
10.2 0.3 3.06 31.212
10.6 0.1 1.06 11.236
sum[x²*p(x)]=9.025+19.208+30+31.212+11.236=100.681
variance of the rate of return=V(x)=sum(x²*p(x))-[sum(x*p(x))]²
variance of the rate of return=V(x)=100.681-(10.03)²
variance of the rate of return=V(x)=100.681-100.6009
variance of the rate of return=V(x)=0.0801