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3 years ago

A car travels at 65 miles per hour write an equation that compares the time with the distance

Mathematics

1 answer:

vichka [17]3 years ago

7 0

Answer:

d = 65t

Step-by-step explanation:

Distance in miles = 65 * the amount of hours

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F(x)=-x^(4)-9x^(3)-24x^(2)-16x what is the relative maxium and minimum?

stepladder [879]

We'll need to find the 1st and 2nd derivatives of F(x) to answer that question.

F '(x) = -4x^3 - 27x^2 - 48x - 16 You must set this = to 0 and solve for the
roots (which we call "critical values).

F "(x) = -12x^2 - 54x - 48

Now suppose you've found the 3 critical values. We use the 2nd derivative to determine which of these is associated with a max or min of the function F(x).

Just supposing that 4 were a critical value, we ask whether or not we have a max or min of F(x) there:

F "(x) = -12x^2 - 54x - 48 becomes F "(4) = -12(4)^2 - 54(4)
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Because F "(4) is negative, the graph of the given
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3 years ago

How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :
</span></span>
$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

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