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Daniel [21]
4 years ago
11

Please help me please! Also, please show your work image shown below in attachment

Mathematics
1 answer:
WINSTONCH [101]4 years ago
6 0

Answer:

see below

Step-by-step explanation:

some of your answers that you currently have are wrong, I'll note those mistakes below

- when factoring (ie 5x+15) only factor out things that can divide both numbers into a whole number ratio

5x+15 = 5(x+3), not (x+3)(x+5)

ie  \frac{10x^2+20x}{x+2}

we see that 10x can divide the numerator in a whole number ratio

10x^2+20x = 10x(x+2), not (x+2)(x+10)

second mistake: the first binomial expansion is incorrect.

you have the expansion formula right, but you added terms wrong, go look at it again

3. x^2+3x+2/ x^2+5x+6

(x+1)(x+2)/(x+3)(x+2)

(x+1)/(x+3)

4. (x^2+6x+8)/(x^2-16)

(x+4)(x+2)/(x+4)(x-4)

(x+2)/(x-4)

5. we can't simplify that any more, x and y are different variables so therefore we cannot cross out stuff on numerator and denominator

6. (x^4y^6)^2

(x^4y^6)(x^4y^6) = x^8y^{12}

remember that (x^a)(x^b) = x^(a+b)

or remember that (x^a)^b = x^ab

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3 years ago
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Answer? I need it for math.
ladessa [460]

Answer:

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3 years ago
Let's assume a typical New Zealand farmer raising sheep owns 250 hectares and averages about 10 livestock units per hectare. If
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Answer:

$50,000

Step-by-step explanation:

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4 0
3 years ago
Perform a first derivative test on the function ​f(x)equals2 x cubed plus 3 x squared minus 120 x plus 6​; ​[minus5​,8​]. Bold
maria [59]

Answer:

a) Critical points

x = 4 and x = -5

b) x = 4 corresponds to a minimum point for the function f(x)

x = - 5 corresponds to a maximum point for the function f(x)

c) The minimum value of f(x) in the interval = -298

The maximum value of f(x) in the interval = 431;

Step-by-step explanation:

f(x) = 2x³ + 3x² - 120x + 6 in the interval [-5, 8]

a) To obtain the critical points, we need to obtain the first derivative of the function with respect to x. Because at critical points of a function, f(x), (df/dx) = 0

f'(x) = (df/dx) = 6x² + 6x - 120 = 0

6x² + 6x - 120 = 0

Solving the quadratic equation,

x = 4 or x = -5

The two critical points, x = 4 and x = -5 are in the interval given [-5, 8] (the interval includes -5 and 8, because it's a closed interval)

b) To investigate the nature of the critical points, we obtain f"(x)

Because f'(x) changes sign at the critical points

If f"(x) > 0, then it's a minimum point and if f"(x) < 0 at c, then it's a maximum point.

f"(x) = (d²f/dx²) = 12x + 6

at critical point x = 4

f"(x) = 12x + 6 = 12(4) + 6 = 54 > 0, hence, x = 4 corresponds to a minimum point.

at critical point x = -5

f"(x) = 12x + 4 = 12(-5) + 4 = -56 < 0, hence, x = -5 corresponds to a maximum point.

c) At x = 4,

f(x) = 2x³ + 3x² - 120x + 6 = 2(4)³ + 3(4)² - 120(4) + 6 = -298

At x = - 5

f(x) = 2x³ + 3x² - 120x + 6 = 2(-5)³ + 3(-5)² - 120(-5) + 6 = 431

7 0
3 years ago
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