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LekaFEV [45]
2 years ago
6

Please help me solve for x

Mathematics
1 answer:
nadya68 [22]2 years ago
4 0

Answer:

x = 4

Step-by-step explanation:

The opposite sides of a parallelogram are congruent, thus

7x - 4 = 24 ( add 4 to both sides )

7x = 28 ( divide both sides by 7 )

x = 4

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3 years ago
How do you do this problem 1/4 +4x/5=11/20?
Nutka1998 [239]
\frac{1}{4}+\frac{4}{5}x=\frac{11}{20}\\\\\frac{4}{5}x=\frac{11}{20}-\frac{1}{4}\\\\\frac{4}{5}x=\frac{11}{20}-\frac{5}{20}\\\\\frac{4}{5}x=\frac{6}{20}\ \ \ \ \ \ \ |\cdot\frac{5}{4}\\\\x=\frac{30}{80}=\frac{3}{8}
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3 years ago
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8 0
3 years ago
A circle has a sector with area 33 pi and a central angle of 11/6 pi radians. What is the area of a circle?
Mashutka [201]

Answer:

36π

Step-by-step explanation:

The area of a circle is given as:

A = \pi r^2

where r = radius of the circle

The area of a sector of a circle is given as:

A_s = \frac{\alpha }{2\pi} * \pi r^2

where α = central angle in radians

Since \pi r^2 is the area of a circle, A, this implies that:

A_s = \frac{\alpha }{360}  * A

A circle has a sector with area 33 pi and a central angle of 11/6 pi radians.

Therefore, the area of the circle, A, is:

33 \pi = \frac{\frac{11 \pi}{6} }{2 \pi} * A\\\\33\pi = \frac{11}{12} * A\\\\=> A = \frac{33\pi * 12}{11}\\ \\A = \frac{396 \pi}{11} \\\\A = 36\pi

The area of the circle is 36π.

5 0
3 years ago
Read 2 more answers
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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