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galben [10]
3 years ago
5

Which classifications apply to the polygon? Check all that apply.

Mathematics
2 answers:
xeze [42]3 years ago
6 0
It’s concave and hexagon
Flura [38]3 years ago
6 0

Answer:

B and D

Step-by-step explanation:

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Help due in 20 mins will mark brainliest.
ValentinkaMS [17]

Answer:

1. The ratio in the table above is a 2.5 : 1 ratio.

2. The second row represents that if you use 15 cups of flour you need 6 tablespoons of flour. (I’m not 100% sure this one is right)

3. 10 cups of flour and 4 tablespoons vanilla

Step-by-step explanation:

3. If you divide your number of flour by 2.5 you get the amount of vanilla you must use

3 0
3 years ago
What is the solution to this inequality?
Alika [10]

Answer:

<h2>A</h2>

Step-by-step explanation:

<em><u>e5uiitweyutwwsuoye34t</u></em>

3 0
3 years ago
Read 2 more answers
Describe the sampling distribution of p(hat). Assume the size of the population is 30,000.
Naya [18.7K]

Answer:

a) \mathbf{\mu_ \hat p = 0.6}

b) \mathbf{\sigma_p =0.01732}

Step-by-step explanation:

Given that:

population mean \mu = 30,000

sample size n = 800

population proportion p = 0.6

a)

The mean of the the sampling distribution is equal to the population proportion.

\mu_ \hat p =  p

\mathbf{\mu_ \hat p = 0.6}

b)

The standard deviation of the sampling distribution can be estimated by using the formula:

\sigma_p = \sqrt{\dfrac{p(1-p)}{n}}

\sigma_p = \sqrt{\dfrac{0.6(1-0.6)}{800}}

\sigma_p = \sqrt{\dfrac{0.6(0.4)}{800}}

\sigma_p = \sqrt{\dfrac{0.24}{800}}

\sigma_p = \sqrt{3 \times 10^{-4}}

\mathbf{\sigma_p =0.01732}

7 0
3 years ago
35,000;3,500;350; I have to find the rule to this pattern
Vikki [24]
The answer is was he number is being divided by 10
3 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
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