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Alecsey [184]
1 year ago
12

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x?

Mathematics
1 answer:
Tema [17]1 year ago
4 0
Your x values are 1.24 and -0.404.
First you need to make the equation equal 0, and you can do this simply by subtracting 5x, so you get
6x² - 3 - 5x = 0
The quadratic formula is (-b +- √b² - 4ac)/2a, where a is the x², b is the x, and c is the value. This means we can just substitute it in.
You find the value of the part inside the square root, which is -5² - 4 × 6 × -3 = 97. Now we can use this to substitute in to (5 +- √97)/12. We can do it with the plus sign, and get 1.24, and then with the subtract sign and get -0.404.

I hope this helps!
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Example:
1+2=3
2+1+3

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8 0
1 year ago
The following table shows the number of hours some teachers in two schools expect students to spend on homework each week: Schoo
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Answer:

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Step-by-step explanation:

Part A

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set in four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

Interquartile range of the data is

IQR=Q_3-Q_1=16-6.5=9.5

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

Interquartile range of the data is

IQR=Q_3-Q_1=15.5-8=7.5

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.

4 0
2 years ago
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7 0
2 years ago
Read 2 more answers
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Answer:

x=15

Step-by-step explanation:

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then subtract 5x from both sides and you have -3x=-45

then finally divide -3 from -45 to get x=15

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4 0
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