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3 years ago

14

Malloy solved the equation −5x − 16 = 8; his work is shown below. Identify the error and where it was made. −5x − 16 = 8 Step 1:

−5x − 16 + 16 = 8 + 16 Step 2: −5x = 24 Step 3: negative five times x over five = twenty-four over five Step 4: x = twenty-four over five

2 answers:

meriva3 years ago

8 0

He did multiplication instead of division

Oliga [24]3 years ago

5 0

Malloy error was in step 3, Malloy put -5x over 5 = 24 over 5, but the correct way to do step 3 is -5x over -5 = 24 over -5 which then x would equal to 24 over -5 or x = -4.8 in simpler terms

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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

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blagie [28]

Answer:

4(2x) - 8x + 4

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2 years ago

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HACTEHA [7]

Answer:

The degree of this monomial is 2.

Step-by-step explanation:

This is because the degree of a monomial (or a polynomial, binomial) is equal to the degree (exponent) of the term in the monomial. In polynomials, it would be the degree of the term with the HIGHEST degree.

-1.75k^2

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3 years ago

Solve for x: 3x − 2 > 5x + 10. (5 points)

velikii [3]

Answer:

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Step-by-step explanation:

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2 years ago