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valentinak56 [21]
3 years ago
14

30x squared-3x-6 what is the answer

Mathematics
1 answer:
Aloiza [94]3 years ago
5 0
The parabola has a vertex of (0.05, -6.075) and goes through the points. (0,-6), (0.5, 0) and (-0.4, 0)
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Simply 3pq+5pr-2qr+qp-6rp​
Agata [3.3K]

Answer:

The answer is 4pq−pr−2qr

5 0
3 years ago
I’m being timed please help guys
tankabanditka [31]

x = 54

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3 0
3 years ago
I need help with algebra
aalyn [17]
To find the equation of a line using two coordinate points, use the slope intercept formula, (y2-y1)/(x2-x1). Using this formula, plug in your values from the two coordinates (-5, -5) and (1, -3). This would look like this:
(-3-(-5))/(1-(-5))
When simplified, you should get 2/6 or 1/3 as your slope or m value. Now that you have your slope, now use your slope intercept equation y=mx+b and plug in your known values. Since you know the m value is 1/3 and you can use one of your set of points to plug into the x and y values, you can plug in these values and solve for b (the coordinate you use doesn't matter but I'm just using the (1,-3)). This would look like this:
-3=1/3(1)+b

When you simplify, you should get -3=1/3+b and when you subtract both sides by 1/3, you should get b=-10/3.

Now that you have your m and b values, you can conclude that your equation for the line would be y=1/3x-10/3



7 0
4 years ago
1
aleksklad [387]

Answer:

.4 is the answer

Step-by-step explanation:

2÷5= .4

7 0
3 years ago
Read 2 more answers
in the fall semester of 2003, the average graduate management admission test (GMAT) of the students at UTC was 500 with a standa
MAVERICK [17]

Answer:

The 2003 scores have a higher coefficient of variation, so they are more dispersed

Step-by-step explanation:

We have to find the coefficient of variation(CV) for both these tests.

CV = \frac{\sigma}{\mu}

In which \mu is the mean and \sigma is the standard deviation.

Whoever has the higher CV has the higher variation, that is, is more dispersed.

2003

\mu = 500, \sigma = 80

CV = \frac{\sigma}{\mu} = \frac{80}{500} = 0.16

2004

\mu = 560, \sigma = 84

CV = \frac{\sigma}{\mu} = \frac{84}{560} = 0.15

The 2003 scores have a higher coefficient of variation, so they are more dispersed.

3 0
4 years ago
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