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damaskus [11]
3 years ago
11

Find the maximum value of c=6x+2y

Mathematics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

  ∞

Step-by-step explanation:

c can have any value you like.

There is no maximum. We say it can approach infinity.

__

<em>Additional comment</em>

There may be some maximum imposed by constraints not shown here. Since we don't know what those constraints are, we cannot tell you what the maximum is.

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10x + (-8)= -3x + 2<br> How do I solve this in pemda order
larisa [96]

Answer:

x= 10/13

Step-by-step explanation:

First, we need to put the variable on one side and the constants on another side.

Also a positive and a negative equal a negative. [this is related to +(-8)]

10x+(-8)= -3x+2

+3x         +3x

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Finally you divide on both sides

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4 0
3 years ago
Which of the following numerical data sets has a mean of 40?
34kurt

Answer:

The answer is B) 21, 34, 65

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21 + 34 + 65 = 120

120 divided by 3 = 40.

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6 0
3 years ago
Solve the equation for all real solutions in simplest form.<br> z^2 - 12z +9= -3
Mars2501 [29]

Answer:

Solving the equation for all real solutions in simplest form.

z^2 - 12z +9= -3 we get \mathbf{z=10.9\: or\: z=1.1}

Step-by-step explanation:

We need to solve the equation for all real solutions in simplest form.

z^2 - 12z +9= -3

First simplifying the equation:

z^2 - 12z +9+3= -3+3\\z^2 - 12z +12= 0

Now, we can solve the equation using quadratic formula:

z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

we have a = 1, b=-12 and c=12

Putting values in formula and finding values of x

z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\z=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\z=\frac{12\pm\sqrt{144-48}}{2}\\z=\frac{12\pm\sqrt{96}}{2}\\z=\frac{12\pm9.8}{2}\\z=\frac{12+9.8}{2}\:or\:z=\frac{12-9.8}{2}\\z=10.9\:or\:z=1.1

So, we get value of z: z=10.9 or z=1.1

Solving the equation for all real solutions in simplest form.

z^2 - 12z +9= -3 we get \mathbf{z=10.9\: or\: z=1.1}

3 0
3 years ago
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