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snow_lady [41]
3 years ago
12

Write an inequality for each sentence. More than 3400 people attended the tea market.

Mathematics
2 answers:
vladimir1956 [14]3 years ago
8 0
The answer would be

~X > 3400~

Hope this helps

Have a great day/night

Feel free to ask any questions
Alla [95]3 years ago
7 0

Answer:

the take video g g go ytffdee

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Need the answer to this question pls help
notsponge [240]

Answer:

12√x^2.y^9

Step-by-step explanation:

x^1/6 * y^3/4

xy ^ (1/6+3/4)

xy ^ (2/12+9/12)

12√x^2 * y^9

4 0
2 years ago
I need help with this question please
Gemiola [76]

Answer:

X = 13

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The functions f, g and h are as follows:
Basile [38]

Answer:

wik2k2k2kw

Step-by-step explanation:

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e

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4 0
3 years ago
Select the correct comparison.
Delicious77 [7]

The true comparison is the typical value is greater in set A. The spread is greater in set B.

<h3>What is the true comparison?</h3>

Spread is used to measure the variability of a data set. Range can be used to measure spread. Range is the difference between the largest number and the smallest number in the dataset.

  • Spread of set A =  7 - 5 = 2
  • Spread of set B = 8 - 1 = 7

The median can be used to measure the typical value of the dataset. Median is the number at the center of the data set.

  • Median of set A = 6
  • Median of set B = 4

To learn more about median, please check: brainly.com/question/14746682

#SPJ1

8 0
2 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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