We have that the logarithmic function has a domain from 0 (without including 0) to plus infinity. That is because of the property that e=2.71 raised to any power is positive. We also have that the square root function is defined only from 0 to plus infinity, since every square of a number is positve or 0. The answer is that the domains differ; they are almost equal, except for the fact that the domain of the square root function additionally contains 0.
Answer:
Step-by-step explanation:
<u>Given equation:</u>
To solve the equation, the L.H.S must be equivalent to the R.H.S. To prove that L.H.S is equivalent to R.H.S, we have to know the value of "y".
The value of the y variable can be found by isolating the variable from its coefficient. In this equation, the y-variable can be isolated by dividing both sides by four. This cancels the coefficient, isolating the y-variable on one side of the equation.
- ⇒ 4y = 12
- ⇒ 4y/4 = 12/4 (Dividing both sides by 4)
Once the y-variable has been isolated, we can simplify the other side of the variable. The expression/variable/number obtained, on the other side of the equation, is the value of the y-variable.
- ⇒ y = 12/4 (Isolating the y-variable)
- ⇒ y = 3 (Simplifying the R.H.S)
Therefore, the value of y is 3.
Learn more about equations: brainly.com/question/27688114
Answer:
I/ (pr)=t
Step-by-step explanation:
I=prt
Divide each side by pr
I/ (pr)=prt/(pr)
I/ (pr)=t
Answer:
72 mins
Step-by-step explanation:
12 mins = 4 tortillas
1 tortilla = 3 mins
3x24=72
72 mins
<h3><em>In AP form 2nd term - 1st term = 3rd term - 2nd term
</em></h3><h3><em>b²-a² = c²-b²
</em></h3><h3><em>b²+b² = c²+a²
</em></h3><h3><em>2b² = c²+a²
</em></h3><h3><em>
</em></h3><h3><em>Add 2ab+2ac+2bc on both sides
</em></h3><h3><em>
</em></h3><h3><em>2b²+2ab+2ac+2bc = a²+c²+ac+ac+bc+bc+ab+ab
</em></h3><h3><em>2b²+2ab+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac
</em></h3><h3><em>2b²+2ab+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac
</em></h3><h3><em>2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)
</em></h3><h3><em>2((ba+b²)+(ca+cb)) = ((ca+cb)+(a²+ab)) + ((cb+c²)+(ab+ac))
</em></h3><h3><em>2(b(a+b)+c(a+b)) = (c(a+b)+a(a+b)) + (c(b+c)+a(b+c)) </em></h3><h3><em>2(b+c)(a+b) = (c+a)(a+b) + (c+a)(b+c)
</em></h3><h3><em>
</em></h3><h3><em>Divide whole by (a+b)(b+c)(c+a)</em></h3><h3><em></em></h3><h3><em>2/c+a = 1/b+c + 1/a+b</em></h3><h3><em>1/c+a + 1/c+a = 1/b+c + 1/a+b</em></h3><h3><em>1/c+a - 1/b+c = 1/a+b - 1/c+a</em></h3><h3><em></em></h3><h3><em>2nd term - 1st term = 3rd term - 2nd term
</em></h3><h3><em>Thus 1/b+c, 1/c+a, 1/a+b are in AP.</em></h3><h3><em></em></h3><h3><em>HOPE IT HELPS !!!</em></h3><h3><em>THANK YOU !!!</em></h3>
