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Tema [17]
3 years ago
7

Find the area of the irregular figure. 8 in. 6 in. A = [ ? Jin.? 5 in. + 10 in. 4 in 4 in.​

Mathematics
2 answers:
kotegsom [21]3 years ago
8 0

I got 108 for the answer

mote1985 [20]3 years ago
6 0
We can divide into 3 shapes
Area of 1st rectangle: 8*6 = 48 in^2
Area of 2nd rectangle: 10*4 = 40 in^2
Area of 3rd rectangle: 5*4 = 20 in^2

Total area = 48 + 40 + 20 = 108 in^2
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Each of the options below describes a relationship that you need to graph in a coordinate plane. which of the options will have
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What is 4•(5+3)=4•8=
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32

Step-by-step explanation:

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A lottery has 60 numbers. To win the jackpot one needs to match all 7 numbers that are drawn by the machine. Is this a PERMUTATI
Vadim26 [7]

Answer: Hence, our required probability is \dfrac{1}{386206920}

Step-by-step explanation:

Since we have given that

Numbers in a lottery = 60

Numbers to win the jackpot = 7 numbers

We need to find the probability to hit the jackpot:

So, our required probability is given by

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Hence, our required probability is [tex]\dfrac{1}{386206920}[/tex

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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
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