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2 years ago

Mike is three years older than his wife Karen.

Mathematics

1 answer:

kkurt [141]2 years ago

8 0

Answer:

1. Karen= g^5 (not sure)
2. Mike = g^5 + 3 (not sure)

3. Karen = 25

Mike = 28

4. Karen = 35

Mike = 38

Step-by-step explanation:

1. if george age was g

karen age= g x 5 = g5

2. mike age= gx5 + 3

3. g = 5

karen age= 5 x 5 = 25

mike 25 + 3

4. g = 7

karen = 7x5 = 35

mike = 35 + 3

(i dont know if this correct or not but it my opinion)

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Answer:

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Step-by-step explanation:

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3 years ago

2x^2+3x=(2x-1)(x+1) Pls do explain

Serggg [28]

<h3>Answer:</h3>

$x=-\frac{1}{2}$

<h3>Step-by-step explanation:</h3>

You could solve this problem by either: Factoring or by using the Quadratic Formula.

SOLVING BY FACTORING STEPS

STEP 1: Move (2x−1)(x+1) to the left side of the equation by subtracting it from both sides.

$2x^2+3x-(2x-1)(x+1)=0$

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$2x+1=0$

STEP 3: Simplify each term.

$2x^2+3x-2x^2-x+1=0$

STEP 4: Simplify by adding terms.

$2x+1=0$

STEP 5: Subtract 1 from both sides of the equation.

$2x=-1$

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SOLVE BY USING THE QUADRATIC FORMULA STEPS

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3 years ago

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3 years ago

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2 years ago

Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

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Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

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Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

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3 years ago