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LuckyWell [14K]
3 years ago
12

Easy please help asap making brainliest.

Mathematics
1 answer:
AfilCa [17]3 years ago
7 0

Answer:

B.

Step-by-step explanation:

When I read a question and I understand I have to gather the information in order to come to an answer.

I hope this helps :)

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Find the probability.
sasho [114]

Answer:

excellent this is ur correcoting answer

8 0
3 years ago
Consider the function below, which has a relative minimum located at (-3 , -18) and a relative maximum located at (1/3, 14/17).
leonid [27]
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27).  But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing.  The first derivative of the function is f'(x)=-3x^2-8x+3.  Set the derivative equal to 0 and factor to find the critical numbers. 0=-3x^2-8x+3, so x = -3 and x = 1/3.  We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing.  From there we will look at our points to determine which fall into the "decreasing" category.  Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞.  In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval.  In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval.  In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval.  In order to determine where our points in question fall, look to the x value.  The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12).  The point (-3, -18) is already a min value.
8 0
3 years ago
1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
                                                     0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
                             x = 0
5 0
3 years ago
Find the maximum and minimum values by evaluating the equation
dezoksy [38]

Answer:

min = -9

max =3

Step-by-step explanation:

C = x-3y

x ≥0

x≤3

y≥0

y≤3

The minimum will be be when x is smallest and y is at its max

x =0 and y = 3

C = 0 - 3(3)

C = 0-9 = -9

The minimum is -9

The maximum occurs when x is largest and y is smallest

x =3 and y = 0

C = 3 - 3(0)

C = 3-0 = 3

The max is 3

7 0
3 years ago
Simplify the expression 2(x - 5)
timofeeve [1]

Answer:

brainliest plsss

Step-by-step explanation:

2(x−5)

=(2)(x+−5)

=(2)(x)+(2)(−5)

=2x−10

7 0
3 years ago
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