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Leviafan [203]
3 years ago
5

HELP I have this on my page for 100 points but no one is looking at it help pls

Mathematics
1 answer:
Vikentia [17]3 years ago
6 0

3 + 4x > -5 = x > -2

First, subtract 3 from both sides of the equation.

3 - 3 + 4x > -5 - 3

= 4x > -8

Now, divide 4 from both sides of the equation.

= x > -2

5(2 - b) > -3(b - 3) = b < 1/2

First multiply the coefficient by the numbers/variables in the parentheses.

10 - 5b > -3b + 9

Now, add 3b to both sides and subtract 10 from both sides of the equation.

-2b > -1

Now divide -1 by -2, but in an inequality, when you divide by a negative number, the sign gets flipped. So, > will turn into <.

b < 1/2

-1.5a + 8 ≥ 17 = a ≤ 6

First, subtract 8 from both sides of the equation.

-1.5a ≥ 9

Now divide 9 by -1.5, but, again, because the number is negative, the sign gets flipped.

a ≤ 6

5w/3 ≥ 6w/4 + 11/2 = w ≥ 22

First, you need to find a common denominator or LCD. In this case, the LCD is 12. So multiply all the fractions by 12 and eliminate factors of 12.

12(5w/3) ≥ 12(6w/4) + 12(11/2)

4(5w) ≥ 3(6w) + 6(11)

20w ≥ 18w + 66

2w ≥ 66

w ≥ 22

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Troyanec [42]

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Step-by-step explanation:

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3 years ago
If a polynomial function f(x) has roots –8, 1, and 6i, what must also be a root of f(x)? A. –6
anyanavicka [17]

Answer:

-6i

Step-by-step explanation:

Complex roots always come in pairs, and those pairs are made up of a positive and a negative version. If 6i is a root, then its negative value, -6i, is also a root.

If you want to know the reasoning, it's along these lines: to even get a complex/imaginary root, we take the square root of a negative value. When you take the square root of any value, your answer is always "plus or minus" whatever the value is. The same thing holds for complex roots. In this case, the polynomial function likely factored to f(x) = (x+8)(x-1)(x^2+36). To solve that equation, you set every factor equal to zero and solve for the x's.

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