Answer:
Step-by-step explanation:
The equation \(8e^x - 1 = 0\) can be solved to find the value of \(x\).
To solve for \(x\), we need to isolate the exponential term, \(e^x\).
Here are the steps to solve the equation:
1. Add 1 to both sides of the equation to isolate the exponential term:
\(8e^x = 1\)
2. Divide both sides of the equation by 8:
\(\frac{{8e^x}}{8} = \frac{1}{8}\)
3. Simplify:
\(e^x = \frac{1}{8}\)
4. To solve for \(x\), we can take the natural logarithm (ln) of both sides of the equation:
\(\ln(e^x) = \ln\left(\frac{1}{8}\right)\)
5. Since \(\ln(e^x)\) and \(e^x\) are inverse functions, they cancel each other out:
\(x = \ln\left(\frac{1}{8}\right)\)
6. Use the properties of logarithms to simplify further:
\(x = \ln(1) - \ln(8)\)
7. Simplify:
\(x = -\ln(8)\)
Therefore, the solution for \(x\) in the equation \(8e^x - 1 = 0\) is \(x = -\ln(8)\).
(x + 1)2(2x – 3)
(x + 1)(3x – 2)(5x – 3)
(x + 1)(2x – 3)(3x – 5)
We have to determine the complete factored form of the given polynomial
.
Let x= -1 in the given polynomial.
So,
So, by factor theorem
(x+1) is a factor of the given polynomial.
So, dividing the given polynomial by (x+1), we get quotient as .
So, = (x+1).
=
=
= is the completely factored form of the given polynomial.
Option D is the correct answer.
Answer:
D) (x + 1)(2x – 3)(3x – 5)
Step-by-step explanation:
Answer:
One solution to this question could be that g(xl=x^2 and f(x)=sqrt{x+1}
Step-by-step explanation:
If g(x)=x^2, f(g(x))=f(x^2)=sqrt{x^2+1}
Answer:
y=10x-10
Hope this helped <3
Answer:
Let's define:
A = distance traveled before deploying the landing gear
B = distance traveled after deploying the landing gear.
We must have that the sum of those two distances must be equal to 600 yards.
A + B = 600 yd.
And we know that:
" the plane will travel half the distance with the landing gear than without the landing gear."
Then we have that:
B = A/2.
Now we can replace this last equation in the first one:
A + B = 600yd
A + A/2 = 600yd.
(3/2)*A = 600yd.
A = (2/3)*600yd = 400yd.
Ohama should deploy the landing gear 400 yd into the runway.