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klasskru [66]
3 years ago
11

Select all the points that are on the graph of the function defined by y = 2x + 3

Mathematics
2 answers:
ale4655 [162]3 years ago
4 0

Answer:

b, d, e, g are the points

Kaylis [27]3 years ago
3 0
All you need to do to solve this is plug in x and y values.

a) -8 + 3 = -5, not 3
b) -2 + 3 = 1
c) 0 + 3 = 3, not 0
d) 4 + 3 = 7
e) 6 + 3 = 9
f) 6 + 3 = 9, not 12
g) 9 + 3 = 12

The correct answers are B, D, E, G
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How many times bigger is 2.7 than 0.03
OlgaM077 [116]
90 times bigger :) hope this helps !
4 0
2 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

4 0
3 years ago
Will enyone help me answer this question please
SSSSS [86.1K]
You could use 4(division) sign 25 + 3 (division sign) 15 * 10, ok?
6 0
3 years ago
A monkey is swinging from a tree. On each swing, she travels along an arc that is 75% as long as the previous swing's arc. The t
mr_godi [17]

Answer:

Total length of the first swing=64 m

Step-by-step explanation:

The total length of all four swings can be expressed as;

Total length of all 4 swings=Length of first swing+length of second swing+length of third swing+length of fourth swing

where;

Total length of all 4 swings=175 m

Length of first swing=x

Length of second swing=75% of length of first swing=(75/100)×x=0.75 x

Length of third swing=75% of length of second swing

Length of third swing=(75/100)×0.75 x=0.5625 x

Length of fourth swing=75% of length of third swing

Length of fourth swing=(75/100)×0.5625 x=0.421875 x

replacing;

Total length of all 4 swings

175=x+0.75 x+0.5625 x+0.421875 x

2.734375 x=175

x=175/2.734375

x=64

Total length of the first swing=x=64 m

3 0
3 years ago
Read 2 more answers
The scores on a test given to all juniors in a school district are normally distributed with a mean of 72 and a standard deviati
faust18 [17]
30% is the correct  answer.
8 0
3 years ago
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