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Natasha_Volkova [10]
2 years ago
11

the formula for the area of a square s to the 2nd power, where s is the side length of the square. What is the area of a square

with a side length of 6 centimeters? Do not include units in your answer
Mathematics
1 answer:
Helen [10]2 years ago
5 0

The answer to this is 36.

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Help me simplify this because i have no idea how to do it and no teacher to help me
VARVARA [1.3K]

Answer:

4sqrt6

Step-by-step explanation:

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2 years ago
What is the percent of change from 5,000 to 900 ?
nadezda [96]

Answer: 82% decrease

Step-by-step explanation:

Initial number =5000

Final number = 900

Change in numbers = 5000 - 900 = 4100

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7 0
3 years ago
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What is the answer to 847x487
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2 years ago
Help would be truly appreciated. Write the polynomial in standard form from the given zeroes of lest degree that has rational co
pishuonlain [190]

Answer:

1) The polynomial in standard form is f(x) = x³ - 9·x² + 23·x - 15

2) The polynomial in standard form is f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The polynomial in standard form is f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The polynomial in standard form is f(x) = x² - (5 + √5)·x + 6 + 3·√5

Step-by-step explanation:

Given that that the polynomial is of least degree, we have;

The standard form is the form, f(x) = a·xⁿ + b·xⁿ⁻¹ +...+ c

1) The zeros of the polynomial are x = 5, 3, 1

Which gives the polynomial in factored form as_f(x) = (x - 5)·(x - 3)·(x - 1)

From which we have;

f(x) = (x - 5)·(x - 3)·(x - 1) = (x - 5)·(x² - 4·x + 3) = x³ - 4·x² + 3·x - 5·x² + 20·x - 15

f(x) = x³ - 9·x² + 23·x - 15

The polynomial in standard form is therefore f(x) = x³ - 9·x² + 23·x - 15

2) The zeros of the polynomial are x = 2, 0, 2·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x)·(x - 2·i)

From which we have;

f(x) = (x - 2)·x·(x - 2·i) = (x² - 2·x)·(x - 2·i) = x³ - 2·i·x² + 2·x² + 4·ix

The polynomial in standard form is therefore f(x) = x³ - (2 - 2·i)·x² + 4·i·x

3) The zeros of the polynomial are x = 2, 1 - 3·i

Which gives the polynomial in factored form as_f(x) = (x - 2)·(x - 1 - 3·i)

From which we have;

f(x) = (x - 2)·(x - (1 - 3·i)) = x² - x + 3·i·x - 2·x + 2 -6·i = x² + 3·i·x - 3·x + 2 - 6·i

The polynomial in standard form is therefore f(x) = x² + (3·i - 3)·x + 2 - 6·i

4) The zeros of the polynomial are x = 3, 2 + √5

Which gives the polynomial in factored form as_f(x) = (x - 3)·(x - (2 + √5))

From which we have;

f(x) = (x - 3)·(x - (2 + √5)) = x² - 2·x - x·√5 - 3·x + 6 + 3·√5

f(x) = x² - 5·x - x·√5 + 6 + 3·√5 = x² - (5 + √5)·x + 6 + 3·√5

The polynomial in standard form is therefore f(x) = x² - (5 + √5)·x + 6 + 3·√5.

5 0
2 years ago
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