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ollegr [7]
3 years ago
6

What is the solution to -4(8 – 3x) >_ 6x – 8?

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

x ≥ 4

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define inequality</u>

-4(8 - 3x) ≥ 6x - 8

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Distribute -4:                              -32 + 12x ≥ 6x - 8
  2. Subtract 6x on both sides:        -32 + 6x ≥ -8
  3. Add 32 on both sides:              6x ≥ 24
  4. Divide 6 on both sides:             x ≥ 4

Here we see that <em>x</em> can be any value greater than or equal to 4.

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What is the equation of the line that passes through the point (6,−4) and has a slope of -1/6
Luda [366]

Answer:

y + 4 = (-1/6)(x - 6)

Step-by-step explanation:

Apply the point-slope formula here:  y - k = m(x - h) becomes

y + 4 = (-1/6)(x - 6)

6 0
3 years ago
Write a system of equations, with one equation describing the cost to bowl at Bowl-o-Rama and the other describing the cost to b
ZanzabumX [31]

Complete question :

Write a system of equations, with one equation describing the cost to bowl at Bowl-o-Rama and the other describing the cost to bowl at Bowling Pinz. For each equation, let x represent the number of games played and let y represent the total cost.

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Bowling Pinz rents shoes for $4 and each game cost $2

Answer:

Bowl-O-Rama:

y = $2 + $2.50x

Bowling pinz:

y = $4 + $2x

Step-by-step explanation:

The total cost is the sum of shoe rental plus the product of the unit game rate and the number of games played.

Total cost, y

x = number of games played

Bowl-O-Rama:

y = Shoe rent + (unit rate per game * number of games)

y = $2 + ($2.50 * x)

y = $2 + $2.50x

Bowling Pinz :

y = Shoe rent + (unit rate per game * number of games)

y = $4 + ($2 * x)

y = $4 + $2x

3 0
2 years ago
If g (x) = 1/x then [g (x+h) - g (x)] /h
lys-0071 [83]

Answer:

\dfrac{-1}{x(x+h)}, h\ne 0

Step-by-step explanation:

If g(x) = \dfrac{1}{x}, then g(x+h) = \dfrac{1}{x+h}. It follows that

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

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Answer:B

Step-by-step explanation:

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