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3x-2y=10 what is ordered pair

\(3x-2y=10\\ \color{blue}y=\frac{3x-10}{2}\)

\(y=1.5x-5\)= 0

\(x_0=\frac{5}{1.5}=\frac{10}{3}\)

\(y=1,5(x-x_0)=1.5(x-3.3\overline 3)\)

x   4     5     6     7

y   1    2.5   4    5.5        \(y=f(x)=1.5(x-3.3\overline3)\)

∑  5    7.5   10  12.5

The equation 3x-2y=10 is satisfied with x = 4

\( 3x-2y=10\\ 3\cdot6-2\cdot4=10\\ 18-8=10\)

  !

I am unsure what this question is asking.

What question do you want answered?

1) What is an ordered pair?

2) What is an example of an ordered pair that satisfies 3x-2y=10?

I have never had problems logging in, either. 

Which calculator are you using? Knowing this will help diagnose the problem and fix it.

Interesting question! I would not consider this a "simple" question. It requires one to think about it!

Before we attempt this problem, we have to understand how the rating system works. Generally, the rating system simply takes the average of all the responses' ratings. Therefore, we can create a formula for it.

\(f(\# \hspace{1mm}\text{of responses})=\frac{x_1+x_2+x_3...x_{\# \hspace{1mm}\text{of responses}}}{\# \hspace{1mm}\text{of responses}}=4.44\)

I will create a table, first:

We have tried the first 10 responses. Will 10 responses ever give an average of 4.44? No. How do I know? We need the integers from 1 to 5 to equal a decimal number. That is impossible. Let's try the first one in the table, 2. 

\(x_1+x_2=8.88\)

If the numbers we can substitute into x are integers, it is impossible to find 2 integers that add up to 8.88 because you cannot add two integers and get a decimal. The same logic can be used for the above ones. Now we must hunt for the least number for n such that 4.44n is an integer. I have effectively changed the question now. I will use a fraction to help me out with this:

But what does this mean? The denominator says the least amount of response, 25. How do I know this? Well, what is 4.44*25=111. 

Now the last question we must ask ourselves is can we make 111 with 25 responses?
 

Yes, we can! You can have 21 5-star responses and 2 2-star responses and 2 1-star responses.

rating of 42/9 

5,5,5,5,5,5,5,5,2  would be one possibility

or

5,5,5,5,5,5,5,4,3

or

lots of other combinations :)

The 9 numbers must add to 42

If it is EXACTLY 4.44 then that is     4 and 11/25

So it would be a minimum of 25 responses.

What answers would you need to give? like 5 5 5 5 4 4 4 4?

In a 5 star rating, how can i achieve a rating of 4.44 in the least amount of responses? The responses can only be 1, 2, 3, 4, or 5.

4.44 .... I thought it was repeater :(

If it was repeater, and yes I know it is not,  then the rating wouldbe 42/9  so you would need a multiple of 9 responses. 

The minimum would be 9

Before attempting to do this problem, you must understand the slope-intercept form of a line. It is the following:
 

\(y=mx+b\)

m = slope of the line

b = the y-intercept (the point where it touches the y-axis)

First, we must find the slope. To do this, we must know another formula. This will solve for m.

\(m=\frac{y_2-y_1}{x_2-x_1}\)

Now that we know the formula, let's plug the given coordinates into the formula above to determine the line's slope:

We have now determined the slope of the line, -7/5. The next step is to figure out the value of b. First, let's look at the equation with the m filled in.

\(y=-\frac{7}{5}x+b\)

Just plug in a coordinate into here and solve for b. It doesn't matter which coordinate you substitute in, either. I'll use the first point, (3,4):

Now that we have both m and b solved, we can write the equation in slope intercept form. 

\(y=mx+b\)

Just replace m and b with the numbers we calculated for both. Therefore, your final answer is:
 

\(y=-\frac{5}{7}x+\frac{41}{5}\) 

With the information given in the diagram, two diagrams can result. Can you tell me which one you mean?

 |\

 |  \

 |    \

 |      \   32in

 |        \

 |          \

 |           \                        

 |------------

       41in

or

        |\

        |  \
        |   \
41in |     \   32in
        |       \
        |         \
        |          \

       ------------

Can you please clarify? When you say "adjacent angle," I am unsure which angle you are referencing. 

Actually, I have noticed another issue, too. Both of my interpretations are impossible triangles because you said the hypotenuse is 32 inches, while one of the legs is 41 inches. This is impossible because the hypotenuse has to be the longest side length.

The circumference of a circle  = 2ñr     where r = the radius.

Suppose the angle of your sector is 30 degrees. Then the length of the arc would be  30/360 X 2ñr.

In general -              arc lenth =       angle of sector/360   X   2ñr.

I will end this controversy by considering both cases.

Case #1: If the side lengths are 1/2 inch.

To solve for this find the volume of the smaller cube. Finding the volume of a cube is actually simple. Just use the following formula. 

\(V=s^3\)

Let V = volume of the cube

Let s = side length

To find how many of cubes with a volume of 1/8in^3 would fit in a 96in^3 rectangular prism, just divide them.

\(\frac{V_{rect.}}{V_{cube}}\)

Let's do that!

Therefore, \(768\) cubes with a length of 1/2in can fit in a rectangular prism with a volume of 96in^3.

Case #2: If the cube has a volume of 1/2in^3

We already know the volume of both cubes, so divide the rectangular prism's volume from the cube's volume:

In this scenario, \(192\) cubes of a volume of 1/2in^3 can fit in a rectangular prism with a volume of 96in^3.

Wow, this question is stirring up a prolonged controversy! I have a major issue with this problem, however.

1. We have no information about these objects.

This means that the combined price of both objects can be \($10000\) or vastly larger. The problem does not give any boundaries to their combined price whatsoever. In this example, this would result in a discount of \(99.99\%\)--not \(100\%\).

Something I can deduce is that the discount must be less than \(100\%\) as a \(100\% \) discount would mean that the item is free. 

Therefore, the maximum discount must be \(99.99\%\leq x<100\%\), or bounded between 99.99% and 100%.

After all of this pondering and contemplating with the current language of the problem, I still feel as if the problem is nonsensical. 

In the form that it is given, \(3x+y=5 \) is in general form.

what is -2 to the 5th power

\((-2)^5=-32\\ and\\ -2^5=-1*2^5=-32\)

using trig

hypotenuse of triangle=(2sqrt3)/sin60 =4

leg of trangle=(2sqrt3)/tan60 =2

trapezoid get 2 triangle

so the perimeter=4*2+2*2+5*2=22

the centroid always divide the median in 2:1 ratio 

so,cp:pm=2:1

cm^2=9+27

cm=6

cp=6(2/3)=4

We have a right triangle, triangle ABC where the legs AB and AC have lengths 6 and 3sqrt3 respectively.

Medians AM and CN meet at point P.

What is the length of CP?

Let \(\vec{A} = \binom{0}{0}\)
Let \(\vec{B} = \binom{6}{0}\)
Let \(\vec{C} = \binom{0}{3\sqrt{3}}\)

\(\mathbf{\vec{P} = \ ?}\)

\(\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 ( \vec{A}+\vec{B}+\vec{C} ) \\ \vec{P} &=& \frac13 \left( \binom{0}{0}+\binom{6}{0}+\binom{0}{3\sqrt{3}} \right) \\ \vec{P} &=& \frac13 \cdot \binom{0+6+0}{0+0+3\sqrt{3} } \\ \vec{P} &=& \frac13 \cdot \binom{6}{3\sqrt{3} } \\ \vec{P} &=& \dbinom{2}{ \sqrt{3} } \\ \hline \end{array}\)

CP = ?

\(\begin{array}{|rcll|} \hline CP &=& |~\vec{C}-\vec{P}~| \\ CP &=& |~\binom{0}{3\sqrt{3}}-\binom{2}{ \sqrt{3} }~| \\ CP &=& |~\binom{0-2}{3\sqrt{3}-\sqrt{3} } ~| \\ CP &=& |~\binom{-2}{2\sqrt{3} } ~| \\ CP &=& \sqrt{(-2)^2+(2\sqrt{3})^2 } \\ CP &=& \sqrt{4+ 4\cdot3 } \\ CP &=& \sqrt{4+ 12 } \\ CP &=& \sqrt{16} \\ \mathbf{ CP } & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

write the eqn into y=mx+b

y=-3x+5 

the slope=m,so the slope=-3 

the slope intercept=b,so the slope int=5

What would be the answer to 6 over the square root of 5 minus the square root of 2?

I think you mean the simplification rather than the answer:)

\(\frac{6}{\sqrt5}-\sqrt 2\\ =\frac{6}{\sqrt5}*\frac{\sqrt5}{\sqrt5}-\sqrt 2\\ =\frac{6\sqrt5}{5}-\frac{5\sqrt 2}{5}\\ =\frac{6\sqrt5-5\sqrt 2}{5}\\\)

OR maybe you meant:

\(\frac{6}{\sqrt5-\sqrt2}\\ =\frac{6}{\sqrt5-\sqrt2}\times \frac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}\\ =\frac{6(\sqrt5+\sqrt2)}{(\sqrt5-\sqrt2)(\sqrt5+\sqrt2)}\\ =\frac{6(\sqrt5+\sqrt2)}{5-2}\\ =2(\sqrt5+\sqrt2)\\\)

This is the formula you would use to find the FV of Julian's deposits:
FV=P{[1 + R]^N - 1/ R}
FV=50 x {[1 + 0.03/12]^(4*12) - 1 / (0.03/12)}
FV=50 x {[1 + 0.0025]^48 - 1 / (0.0025)}
FV=50 x {[1.0025]^48 - 1 / (0.0025)}
FV=50 x         50.93121.......
FV=$2,546.56 - This is what Julian will have at the end of 4 years.

I repeat:

Do you mean that the volume is   0.5cm^3 or do you mean that the side length of the cube is 0.5cm?

THEY ARE DIFFERENT!

BUT GUEST ANSWER  #3   IS CORRECT.

Thanks #3  :)

how to write an equation, in general form, from these descriptions of lines?

a) )a vertical line passing through the point (3,5)

b) a horizontal line passing through the point (-2, 6)

c) the x-axis

d) the y-axis

a)

x=3

b)

y=6

c)

y=0

d)

x=0

  !

We can simplify this expression:

6ab - ba + 3ab + a

                                         ba  =  ab    , so we can write...

6ab - ab + 3ab + a

                                         6 apples - 1 apple + 3 apples  =  8 apples     So...

                                         6ab - 1ab + 3ab  =  8ab

8ab + a