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dem82
3 months ago
14

Given f(x) = x3 – 2x2 – x + 2, the roots of f(x) are

Mathematics
2 answers:
Inessa [12.5K]3 months ago
8 0

Answer:

For Question 1: -1, 1, 2

For Question 2: B

For Question 3: 1, 2, below

Step-by-step explanation:

PIT_PIT [12.4K]3 months ago
3 0
The expression in question is:

f(x) = x³ – 2x² – x + 2

To determine the roots, follow these steps:

1. Set the equation to zero:

0 = x³ – 2x² – x + 2

2. Factor the equation to get:

(x-2)(x-1)(x+1) = 0

3. The roots can be identified as follows:

x1 = -1
x2 = 1
x3 = 2
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Nate has $50 to spend at the grocery store. he fills his shopping cart with items totaling $46 at checkout he willhave to pay 6%
zzz [12365]

Answer:

Yes

Step-by-step explanation:

This is because 46 multiplied by 1.06 equals 48.76, which is under $50.

4 0
2 months ago
Read 2 more answers
The table shows ordered pairs of the function y = 16 + 0.5x .
Leona [12618]

Answer: The unknown values x and y correspond to 8 and 20, specifically;

(x, y) = (8, 20)

Step-by-step explanation: The equation y = 16 + 0.5x represents a linear relationship that can be illustrated with a graph. This indicates that values for x and y can be located at various points on the line.

The ordered pairs signify that for each x value, there exists a matching y value.

The values listed in a two-column format for x and y all fulfill the equation y = 16 + 0.5x. Observing the first example, the pair (-4, 14) is presented.

This reveals that when x is -4, y will be 14.

Where y = 16 + 0.5x

y = 16 + 0.5(-4)

y = 16 - 2

y = 14

Thus, the first pair, similar to the other pairs, satisfies the equation.

Consequently, by reviewing the options provided, we can deduce which one fulfills the equation.

(option 1) If x = 0

y = 16 + 0.5(0)

y = 16 + 0

y = 16

(0, 16)

(option 2) If x = 5

y = 16 + 0.5(5)

y = 16 + 2.5

y = 18.5

(5, 18.5)

(option 3) If x = 8

y = 16 + 0.5(8)

y = 16 + 4

y = 20

(8, 20)

Our calculations confirm that the third option (8, 20) is the correct ordered pair for x and y.

4 0
3 months ago
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