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3 years ago

Please help thanks !!

Mathematics

1 answer:

otez555 [7]3 years ago

6 0

Answer:

x=7

7-4=3

7+2=9

Step-by-step explanation:

Triangle ABE and Triangle CDB are similar

4*3=12

ABE is 3 times bigger than CDB

3(x-4)=x+2

3x-12=x+2

2x-12=2

2x=14

x=7

7-4=3

7+2=9

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What does the relationship between the mean and median reveal about the shape of the data? The mean is less than the median, so

IgorC [24]

Answer:

The mean is equal to the median, so the data is symmetrical

Step-by-step explanation:

Here is the data.

10 5 8 10 12 6

8 10 15 6 12 18

The given data: 10 5 8 10 12 6 8 10 15 6 12 18

For finding the Mean, we will have to add all numbers together and divide it by total number. i.e sum of terms divided by number of terms

Mean= 10+5+8+10+12+6+8+10+15+6+12+18 ÷ 12

Mean = 120 ÷ 12 = 10

For finding the Median, first we need to rearrange the data in ascending order

5 6 6 8 8 10 10 10 12 12 15 18

We can see that the middle values are 10 and 10. So, the median will be the average of those two middle values.

Median = 10+10 ÷ 2

Median = 20 ÷ 2 = 10

From the calculation, we can see that both the median and mean are equal so, the data is symmetrical

5 0

3 years ago

Read 2 more answers

A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

$g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)$

Then, first and second derivatives of the function are, respectively:

First derivative

$g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)$

$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

$g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x$

$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

$g''(x) = 12\cdot x^{2}-30\cdot x -28$

Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$