we'll use the same log cancellation rule, since this is pretty much the same thing as the other, just recall that ln = logₑ.
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x} \\\\\\ \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)+ln(5)=2\implies ln(x\cdot 5)=2\implies log_e(5x)=2 \\\\\\ e^{log_e(5x)}=e^2\implies 5x=e^2\implies x=\cfrac{e^2}{5}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%0A%5C%5C%5C%5C%0Alog_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Blogarithm%20of%20factors%7D%0A%5C%5C%5C%5C%0Alog_a%28xy%29%5Cimplies%20log_a%28x%29%2Blog_a%28y%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29%2Bln%285%29%3D2%5Cimplies%20ln%28x%5Ccdot%205%29%3D2%5Cimplies%20log_e%285x%29%3D2%0A%5C%5C%5C%5C%5C%5C%0Ae%5E%7Blog_e%285x%29%7D%3De%5E2%5Cimplies%205x%3De%5E2%5Cimplies%20x%3D%5Ccfrac%7Be%5E2%7D%7B5%7D%20)
![\bf \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)-ln(5)=2\implies ln\left( \cfrac{x}{5} \right)=2\implies log_e\left( \cfrac{x}{5} \right)=2\implies e^{log_e\left( \frac{x}{5} \right)}=e^2 \\\\\\ \cfrac{x}{5}=e^2\implies x=5e^2](https://tex.z-dn.net/?f=%20%5Cbf%20%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29-ln%285%29%3D2%5Cimplies%20ln%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20log_e%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20e%5E%7Blog_e%5Cleft%28%20%5Cfrac%7Bx%7D%7B5%7D%20%5Cright%29%7D%3De%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%7D%7B5%7D%3De%5E2%5Cimplies%20x%3D5e%5E2%20)
Answer:
c = 169
Step-by-step explanation:
We can use FOIL to determine this after factoring out the equation.
(x+13)(x+13)
F: (x*x)
O: (x*13)
I: (13*x)
L: (13*13)
x^2 + 26x + 169
X = Ishaan
y = Christopher
3y = x
x = y + 14 Put the first equation into the second.
3y = y + 14
2y = 14
y = 7
Christopher is 7
Ishaan is 21
A fisherman can row upstream at 4 mph and downstream at 6 mph. He started rowing upstream until he got tired and then rowed downstream to his starting point. The distance of how far the fisherman row if the entire trip took 2 hours is 9.6 miles
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The speed of an object refers to the change of the object's distance within a specified time.
Mathematically, we can have;

Distance = speed × time
From the information given;
- The fisherman row upstream for 4 miles/ hour.
- The distance covered by the fisherman = 4 × t₁ ----- (1)
- The fisherman row downstream at 6 miles/ hour
- The distance covered by the fisherman = 6 × t₂ ------ (2)
- SInce it took the entire trip 2 hours, we can infer that:
From equation (2), replace the above value of t₂ into equation (2)
- d = 6 t₂
- d = 6(2 - t₁)
- d = 12 - 6t₁
From equation (1)
Equation both distance together, we have:
- 12 - 6t₁ = 4t₁
- 12 = 4t₁ + 6t₁
- 12 = 10t₁
- t₁ = 12/10
- t₁ = 1.2 hours
From t₂ = 2 - t₁
- t₂ = 2 - 1.2
- t₂ = 0.8 hours
From d = 4t₁
Replace t₁ with 1.2 hours
- d = 4(1.2 hours)
- d = 4.8 miles
Also, from d = 6t₂
Therefore, we can conclude that the distance of how far the fisherman row if the entire trip took 2 hours is (4.8+4.8) miles = 9.6 miles
Learn more about distance here:
brainly.com/question/12319416?referrer=searchResults
Answer:
i think its the third one correct me if im wrong 0:)