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VMariaS [17]
1 year ago
12

Without evaluating, order

Mathematics
1 answer:
Iteru [2.4K]1 year ago
8 0

The order of the terms from l<u>east to greates</u>t is 5^-7, 5^0 and 5^4

<h3>Exponents and indices</h3>

Exponents are written as power to a number or expression. According to the question we are to arrange the exponents  5^-7, 5^4 and 5^0 from least to the greatest.

The best approach is to check their exponents and arrange them from least to greatest with evaluating each terms

From the given exponents, the arrangement from least to greatest is -7, 0 and 4. Hence we can conclude that the order of the terms from l<u>east to greates</u>t is 5^-7, 5^0 and 5^4

Learn more on exponent ordering here: brainly.com/question/24468862

#SPJ1

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Scientific notation of 1290000000
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What is the slope of a line that is perpendicular to the line y = 1/6 x + 4? The slope of the line is .
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Answer:

-6

Step-by-step explanation:

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7 0
3 years ago
A quaderlateral has the following angle measurements :100,90,45, and?
Andrews [41]
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7 0
4 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
Can someone help me with this please....ill give points and brainless​
den301095 [7]

Answer:

The total surface area of the given cone is

S = 4( \sqrt{3}+1)\pi  cm^2

Step-by-step explanation:

<u>Step 1</u>:-

<u>Formula:-</u>

The total surface area of cone is

S.A = \pi r l +B

S.A = \pi r l +area of base

S.A = \pi r l+ \pi r^2........(1)

<u>Step 2:-</u>

Given slant height l = 2\sqrt{3} cm

and radius of the circle (base) = 2

substituting these values in equation (1)

S.A = \pi (2)(2\sqrt{3} ) + \pi (4)

on simplification, we get

S.A = \pi (4\sqrt{3}+4)

<u>Final answer</u> :-

S.A = 4 (\sqrt{3}+1)\pi cm^2

8 0
3 years ago
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