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alukav5142 [94]
3 years ago
15

Solve the inequality 4(k − 5) + 12k ≥ −4

Mathematics
2 answers:
Dominik [7]3 years ago
8 0
Isolate the variable by dividing each side by factors that don't contain the variable.
Inequality Form:
k
≥
1
k
≥
1
Interval Notation:
[
1
,
∞
)
[
1
,
∞
)
IgorC [24]3 years ago
5 0

Answer:

k ≥ 1

Step-by-step explanation:

Step 1: Write inequality

4(k - 5) + 12k ≥ -4

Step 2: Solve for <em>k</em>

  1. Distribute 4: 4k - 20 + 12k ≥ -4
  2. Combine like terms: 16k - 20 ≥ -4
  3. Add 20 to both sides: 16k ≥ 16
  4. Divide 16 on both sides: k ≥ 1
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Simplify the rational expressions. State any excluded values. Show your work. (3x - 6)/(x - 2)
cluponka [151]
Hello,

Put the 3 as a common factor.

Look:

3x - 6 = 3 . ( x - 2 )

Then,

( 3x - 6) / (x - 2) = 3 .( x - 2) / (x -2)

Now simplity ( x -2)

= 3
5 0
3 years ago
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Rewrite the equation in standard form y= 4/5x+0.85. Enter answer as ax+by=c or ax+by+c=0
Paraphin [41]

At very minimum, move (4/5)x to the other side of the given equation.  Then:

-(4/5)x + y = 0.85.  This is the equation in standard form.

3 0
3 years ago
The base of a cylindrical candle has an area of
Fudgin [204]

The volume of the cylindrical candle that has a base area of 12.56 inches is 12.56h inches³.

<h3>Volume of of cylinder</h3>

volume of a cylinder = πr²h

where

  • r = radius
  • h = height

Therefore,

volume of a cylinder = πr²h

Area of the base = πr²

Therefore,

Area of the base = 12.56 inches²

volume of the cylindrical candle = 12.56h inches³

learn more on volume here: brainly.com/question/11459769

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7 0
2 years ago
Evaluate<br> 3-2.(4.6). 23<br> Enter your answer in the box.
Taya2010 [7]

Answer:

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Step-by-step explanation:

6 0
3 years ago
The simplified expression
Romashka-Z-Leto [24]

Answer:

5x^2 y^2

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. \sqrt[n]{x^a} =x^{\frac{a}{n}}

2. \sqrt{x}\sqrt{x}  =x

3.  \sqrt{x} \sqrt{y}=\sqrt{x*y}

Area of a triangle is given by  1/2 * base * height, so we do that and simplify:

A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} }  y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2

6 0
3 years ago
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