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3 years ago

15

The Square of a positive number is 12 more than 4 times the number. Find the number. [ Only an algebraic solution will be accept

ed]

1 answer:

Andreas93 [3]3 years ago

6 0

X^2=4x+12
x^2-4x-12=0
(x+2)(x-6)=0
x=6 is the answer

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Define equivalent expressions. Give an example of two equivalent expressions.

Tasya [4]

Answer:

Hello There!

Step-by-step explanation:

Equivalent expressions are expressions that work the same even though they look different.

Examples of two equivalent expressions are:

-4(x + 2) and 4x + 8

-2y+5y−5+8 and 7y+3

These are equivalent expressions as they have the same value.

hope this helps,have a great day!!

~Pinky~

3 0

3 years ago

A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

$g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)$

Then, first and second derivatives of the function are, respectively:

First derivative

$g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)$

$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

$g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x$

$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

$g''(x) = 12\cdot x^{2}-30\cdot x -28$

Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

$g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28$

$g''(0) = -28$

This value leads to a local maximum.

x = 5.118

$g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28$

$g''(5.118) = 132.787$

This value leads to a local minimum.

x = -1.368

$g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28$

$g''(-1.368) = 35.497$

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

$g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}$

$g(0) = 0$

Local minima

$g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}$

$g(5.118) = -350.90$

$g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}$

$g(-1.368) = -9.90$

7 0

3 years ago

K^2+k=0 what does k=

FinnZ [79.3K]

Answer:

k = K^2

Step-by-step explanation:

Subtract K^2 from both sides of the equation.

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3 years ago

Following an experiment, Jamie estimates that his probability of making a three-pointer with a basketball is 0.8. If he shoots 1

dem82 [27]

Multiply the number of attempts with the probability,
160 x 0.8 = 128 attempts made.

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