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GenaCL600 [577]
3 years ago
9

-2.5(x-4)=-3+4 What is the solution to the equation

Mathematics
1 answer:
gavmur [86]3 years ago
6 0

thank me the answer is x=3.6

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I don’t understand these
BaLLatris [955]
So these are basically isolating the variables.
The first equation is 3g + 5 =17.
In order to isolate the variable, we would have to get g by itself, that means 5 would have to go. In order to do this, we would do the opposite. Since it is positive 5 we would add negative 5, in order for it to disappear. This works because a positive 5 and negative 5 cancel each other out. Whatever you do to one side of the equation you have to do to the other, since we subtract 5 on one side we have to subtract 5 on the other. Therefore we would do 17-5.
Now we have 3g=12
We know that 3g is basically 3 multiplied by g. The opposite of multiplication is division.Therefore we would divide by 3 on both sides.
The answer to the first question would be g= 4.
And if you want to check if your answer is correct you plug the value in.
So
3(4) + 5 =17


5 0
3 years ago
Please someone help me please!I'm struggling and I'll give extra point's!
alukav5142 [94]

Answer:

The x variable has an exponent of 2

Step-by-step explanation:

To be a linear equation, the highest power must be 1 on the variables

2x^2 +y =7 is quadratic since x^2 has a power of 2

3 0
3 years ago
When you solve a problem involving money, what can a negative number represent
loris [4]
When you owe money it is a negative because when you earn money you have to pay it back
6 0
3 years ago
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
If you know can you please help me??
Nezavi [6.7K]
0.42 because i said so
4 0
3 years ago
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