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Phantasy [73]
3 years ago
15

The equation -2k+7=-3 solve this equation

Mathematics
2 answers:
Vaselesa [24]3 years ago
6 0

-2k+7=-3

     -7=-3

-2k      -6

Divive but 2k over to 6 so 6/2

I got 3

[Other person for k=5 but I think they are wrong.]

klemol [59]3 years ago
5 0
The correct answer to your question is  k=5 .
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Bella has a stack of 41 bills (ones, fives, and tens) in an envelope. She has 3 less tens than fives and the quantity of ones do
AleksandrR [38]

Answer:

x+(x-3)+(2*x)=41

Step-by-step explanation:

We know that the other two bills are based of of a constant value for fives. Therefore, the fives can be x. There are 3 less tens than fives, so the number of tens equals x-3. We double the number of fives to get the number of ones, so the number of ones equals 2 times x. When we add all 3 of those together, we get our total number of 41. If you want to solve, you have to solve for x, then you can plug it back into the formula. I hope this helps!

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What is −n+(−4)−(−4n)+6
luda_lava [24]

Answer:

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Step-by-step explanation:

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3 years ago
For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7
Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be <em>approaching to a value </em>\infty<em> or it should not give us the </em>\frac{0}{0}<em> (zero by zero) form </em> when the input is given to the function.

The value of function will depend on the denominator.

Now, let us consider the given functions one by one:

1. 5y+2

Here denominator is 1. So, it can not attain a value \infty or \frac{0}{0}<em> (zero by zero) form </em>

So, for all real numbers, the function is defined.

2.\ \dfrac{18}{y}

At y = 0, the value

At\ y =0,  \dfrac{18}{y} \rightarrow \infty

So, the given function is <em>defined for all real numbers except y = 0</em>

<em></em>

<em></em>3.\ \dfrac{1}{x+7}<em></em>

Let us consider denominator:

x + 7 can be zero at a value x = -7

At\ x =-7,  \dfrac{1}{x+7} \rightarrow \infty

So, the given function is <em>defined for all real numbers except x = -7</em>

<em></em>

4.\ \dfrac{2b}{10-b}

Let us consider denominator:

10-b can be zero at a value b = 10

At\ b =10,  \dfrac{2b}{10-b} \rightarrow \infty

So, the given function is <em>defined for all real numbers except b = 10</em>

<em></em>

7 0
3 years ago
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stira [4]

Answer:

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A

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