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julsineya [31]
3 years ago
10

I need help with this algebra 2 question, please dont answer if you dont know

Mathematics
1 answer:
Illusion [34]3 years ago
4 0

Answer:

cubic quadratic linear quadratic. first 4 rest i dunno

Step-by-step explanation:

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A line with a slope of 1 passes through the point (18, 15). What is its equation in slope-intercept form? Write your answer usin
Setler79 [48]

Answer:

y = x - 13

Step-by-step explanation:

Given parameters:

  Slope of the line  =  1

  Coordinate = (18, 15)

Unknown:

Equation of the line in slope-intercept format = ?

Solution:

The equation of a line is expressed as;

         y  =  mx + c

where y and x are the coordinates

          m is the slope

          c is the y-intercept

Now, sine x = 18 and y = 5, let us find c;

        5 = 1(18) + c

      5 = 18 + c

       c = -13;

 The equation of the line is;

       y  = 1(x) + (-13)

       y = x - 13

4 0
2 years ago
How to solve this derivative.<br><br> f(x)={ln(3x)}^2x<br> with steps...
AysviL [449]

Rewrite <em>f(x)</em> as a nested exponential-logarithm expression :

\left(\ln(3x)\right)^{2x} = \exp\left(\ln\left(\ln(3x)\right)^{2x}\right)

(where \exp(x) = e^x)

One of the properties of logarithms lets us drop the exponent as a coefficient:

\exp\left(\ln\left(\ln(3x)\right)^{2x}\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)

Now, by the chain rule, we have

f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'

By the product rule,

f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right)

By the chain rule again,

f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)'  \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)

Then simplify this to

f'(x) = \boxed{\left(\ln(3x)\right)^{2x} \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)}

8 0
3 years ago
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