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2 years ago

The equation of liner is y = -x - 6. Line s includes the point (-1, -3) and is parallel to line

Mathematics

1 answer:

Reptile [31]2 years ago

4 0

Answer:

90 hardcover books

Step-by-step explanation:

We can solve this by setting up a couple of equations.

Let's allow x to represent the number of paperbacks Tim owns, and allow y to represent the number of hardcover books he owns.

Using the information in the question, we can write the equations:

1)x = 4y-3

2) x+y=447

Let's rearrange equation 1 so that it is in standard form:

x-4y=-3

And then let's multiply equation 2 by 4 so that we can cancel out y when we solve the system of equations:

4(x+y=447)

4x+4y=1,788

Then we can add the two equations and solve for x:

1) x-4y=-3

+ 2)4x+4y=1,788

------------------------------------

5x=1,785

x=357

So now we now the number of paperback books Tim has is 357. Let's plug this into one of the original equations to solve for the number of hardcover books (y):

357+y=447

y=90

And now we know that Tim owns 90 hardcover books.

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LenaWriter [7]

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3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$