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DerKrebs [107]
3 years ago
6

Y=e^x how do I solve this

Mathematics
1 answer:
kolezko [41]3 years ago
6 0
If you're trying to solve for x, keep in mind the ln (natural log) of e is 1. So take the natural log of both sides, which will give you ln Y = ln e^x. The ln and e essentially cancel out, so you're left with x = ln Y. 
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elimination was used to solve a system of equations. one of the intermediate steps led to the equation 2x= 8. which of the follo
konstantin123 [22]

9514 1404 393

Answer:

  C

Step-by-step explanation:

The step shown indicates that y was eliminated from the equations. For choices A, B, D, this is done by adding the equations together (the y-coefficients are opposites). The result of doing that gives x-terms of 4x, 0x, and 0x, respectively. These x-terms do not match the one given: 2x.

For choice C, the y-term is eliminated by subtracting twice the second equation from the first. Doing that gives ...

  (4x +2y) -2(x +y) = (14) -2(3)

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8 0
3 years ago
Rewrite the equation in vertex form. then find the vertex of the graph. y=-3x^2-5x+1
amm1812

Answer: y = -3(x + \frac{5}{6})² + \frac{37}{12}, (-\frac{5}{6}, \frac{37}{12})

<u>Step-by-step explanation:</u>

First, you need to complete the square:

y   = -3x² - 5x + 1

<u> -1  </u>   <u>                -1  </u>

y - 1 = -3x² - 5x

y - 1 = -3(x² + \frac{5}{3}x

y - 1 + -3(\frac{25}{36}) = -3(x² + \frac{5}{3}x + \frac{25}{36})

           ↑                     ↓            ↑

                                  \frac{5}{3*2} = (\frac{5}{3*2})^{2}

y - 1 - \frac{25}{12} = -3(x + \frac{5}{6})²

y - \frac{12}{12} - \frac{25}{12} = -3(x + \frac{5}{6})²

y  - \frac{37}{12} = -3(x + \frac{5}{6})²

y = -3(x + \frac{5}{6})² + \frac{37}{12}

Now, it is in the form of y = a(x - h)² + k   <em>where (h, k) is the vertex</em>

Vertex = (-\frac{5}{6}, \frac{37}{12})

8 0
3 years ago
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