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zysi [14]
3 years ago
9

What is the inequality to 4-3x≥7

Mathematics
1 answer:
Stella [2.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

4-3x≥7

4-7≥3x

-3≥3x

or

3x≤-3

x≤-1

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3 years ago
4(5x + 9)^2 – 33 = -21
ozzi

Answer:

x = sqrt(3)/5 - 9/5 or x = -9/5 - sqrt(3)/5

Step-by-step explanation by completing the square:

Solve for x over the real numbers:

4 (5 x + 9)^2 - 33 = -21

Add 33 to both sides:

4 (5 x + 9)^2 = 12

Divide both sides by 4:

(5 x + 9)^2 = 3

Take the square root of both sides:

5 x + 9 = sqrt(3) or 5 x + 9 = -sqrt(3)

Subtract 9 from both sides:

5 x = sqrt(3) - 9 or 5 x + 9 = -sqrt(3)

Divide both sides by 5:

x = sqrt(3)/5 - 9/5 or 5 x + 9 = -sqrt(3)

Subtract 9 from both sides:

x = sqrt(3)/5 - 9/5 or 5 x = -9 - sqrt(3)

Divide both sides by 5:

Answer:  x = sqrt(3)/5 - 9/5 or x = -9/5 - sqrt(3)/5

7 0
3 years ago
Read 2 more answers
We want to find the zeros of this polynomial p(x)= ( x^2-1)(x^2-5x+6)
Scrat [10]

Answer:

see explanation

Step-by-step explanation:

To find the zeros let p(x) = 0 , that is

(x² - 1)(x² - 5x + 6) = 0

Factorise each factor

x² - 1 ← is a difference of squares and factors as (x - 1)(x + 1)

x² - 5x + 6 = (x - 2)(x - 3), thus

(x - 1)(x + 1)(x - 2)(x - 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x + 1 = 0 ⇒ x = - 1

x - 2 = 0 ⇒ x = 2

x - 3 = 0 ⇒ x = 3

The zeros are x = ± 1, x = 2, x = 3

4 0
3 years ago
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