All categories

3 years ago

Follow the process of completing the square to solve 2x2 + 8x - 12 = 0.

1 answer:

8 0

Please: Use "^" to denote exponentiation: <span>2x^2 + 8x - 12 = 0

Reduce this by div. every term by 2: </span><span>x^2 + 4x - 6 = 0

Here a=1, b=4 and c = -6. Square half of b, obtaining (4/2)^2 = 4, and add, and then subtract, this 4 to x^2 + 4x - 6:

</span> x^2 + 4x +4 - 4 - 6 = 0. Rewrite the square as (x+2)^2, obtaining new equation

(x+2)^2 = 10. Take the sqrt of both sides: x+2 = plus or minus sqrt(10).

Finally, solve for x: x = -2 plus or minus sqrt(10).

You might be interested in

Lifetime Fitness charges $90 a month for membership plus $50 per session with a trainer. Which equation can be used to determine

Taya2010 [7]

Answer:

50x / 30/x i think

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Approximately 12.6% of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin

bagirrra123 [75]

Answer:

Step-by-step explanation:

Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit.

a) P(X = n) = q(n-1)p, where q = 1 - p.

From the information given, probability if success, p = 12.6/100 = 0.126

b) for n = 3, the probability value from the geometric probability distribution calculator is

P(n = 3) = 0.096

For n = 5, the probability value from the geometric probability distribution calculator is

P(n = 5) = 0.074

For n = 12, the probability value from the geometric probability distribution calculator is

P(n = 12) = 0.8

c) For n ≥ 5, the probability value from the geometric probability distribution calculator is

P(n ≥ 5) = 0.58

d) the expected number of apples that must be examined to find the first one with bitter pit is the mean.

Mean = 1/p

Mean = 1/0.126 = 7.9

Approximately 8 apples

7 0

3 years ago

Find the domains of the following expressions: 5/the square root of 5-5x

AleksandrR [38]

Interval notation

(-infinity sign, 1)

Set-builder notation

{x|x less than equal to 1}

7 0

3 years ago

A pitcher could hold 2/12 of a gallon of water. If Roger filled up 9 pitchers, how much water would he have?

VashaNatasha [74]

1.5 because 2/12=1/6 9 x (1/6) =1.5 gallons of water.

6 0

3 years ago

A company wishes to manufacture some boxes out of card. The boxes will have 6 sides (i.e. they covered at the top). They wish th

Serhud [2]

Answer:

The dimensions are, base $b=\sqrt[3]{200}$ , depth $d=\sqrt[3]{200}$ and height $h=\sqrt[3]{200}$ .

Step-by-step explanation:

First we have to understand the problem, we have a box of unknown dimensions (base $b$ , depth $d$ and height $h$ ), and we want to optimize the used material in the box. We know the volume $V$ we want, how we want to optimize the card used in the box we need to minimize the Area $A$ of the box.

The equations are then, for Volume

$V=200cm^3 = b.h.d$

For Area

$A=2.b.h+2.d.h+2.b.d$

From the Volume equation we clear the variable $b$ to get,

$b=\frac{200}{d.h}$

And we replace this value into the Area equation to get,

$A=2.(\frac{200}{d.h} ).h+2.d.h+2.(\frac{200}{d.h} ).d$

$A=2.(\frac{200}{d} )+2.d.h+2.(\frac{200}{h} )$

So, we have our function $f(x,y)=A(d,h)$ , which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting

$\frac{\partial A}{\partial d} =-\frac{400}{d^2}+2h=0$

$\frac{\partial A}{\partial h} =-\frac{400}{h^2}+2d=0$

After solving the system of equations, we get that the optimum point value is $d=\sqrt[3]{200}$ and $h=\sqrt[3]{200}$ , replacing this values into the equation of variable $b$ we get $b=\sqrt[3]{200}$ .

Now, we have to check with the hessian matrix if the value is a minimum,

The hessian matrix is defined as,

$H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]$

we know that,

$\frac{\partial^2 A}{\partial d^2}=\frac{\partial}{\partial d}(-\frac{400}{d^2}+2h )=\frac{800}{d^3}$

$\frac{\partial^2 A}{\partial h^2}=\frac{\partial}{\partial h}(-\frac{400}{h^2}+2d )=\frac{800}{h^3}$

$\frac{\partial^2 A}{\partial d \partial h}=\frac{\partial^2 A}{\partial h \partial d}=\frac{\partial}{\partial h}(-\frac{400}{d^2}+2h )=2$

Then, our matrix is

$H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]$

Now, we found the eigenvalues of the matrix as follow

$det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0$

Solving for $\lambda$ , we get that the eigenvalues are: $\lambda_1=2$ and $\lambda_2=6$ , how both are positive the Hessian matrix is positive definite which means that the function $A(d,h)$ is minimum at that point.

4 0

3 years ago