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sammy [17]
3 years ago
8

How do you solve: negative eight is less than or equal to 8x

Mathematics
2 answers:
devlian [24]3 years ago
7 0
-8≤8x
Divide both side by 8
-1≤x
x≤1
(When u multiply or divide by a negetive number direction of  inequality chenges)
stiv31 [10]3 years ago
7 0
Less than because the 8 is a negative anytime u have a negative u subtract positive u add so it's Less than
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Step-by-step explanation:

use 2 nonzero numbers but without the decimal for rounding

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What is the unit rate of 183 miles in three hours
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The legs of a right triangle are 8 and 9 units long. What is the area of the triangle in square units?
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36 Hope this helps Im so sorry if this is wrong.

Step-by-step explanation:

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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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NeX [460]

9514 1404 393

Answer:

  no solution

Step-by-step explanation:

Simplifying the given equation, you have ...

  -6(x -2) +3x = -3(x +3) +21 . . . . . given

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  -3x +12 = -3x +13 . . . . . . . . . . . . collect terms

  0 = 1 . . . . . . . . . . . . . add (3x-12) to both sides

There is no value of x that will make this true.

The equation has NO SOLUTION.

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