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12345 [234]
2 years ago
11

I need some help please :)

Mathematics
2 answers:
Butoxors [25]2 years ago
6 0

Answer:

c

Step-by-step explanation:

kupik [55]2 years ago
5 0
A a a a a a a a a a a a
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Solve for x<br><br> 1/3x=7/6 + 1/9
zimovet [89]

Answer:

x=26/3 because you need to comine like terms as usual

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
I how do I do this ?!??
stiv31 [10]

Answer:

m = -3/2 and b = 4

Step-by-step explanation:

m = slope and u go down 3 and over 2 so its -3/2

b = your y intercept and its intercepting through 4 on the y-axis

5 0
3 years ago
Three numbers are in the ratio 2:5:7 and their lcm is 490 find the square root of the largest number
ohaa [14]

Answer:

49

Step-by-step explanation:

The ratio of three numbers = 2:5:7

Then the numbers are 2x,5x and 7x

Given that the LCM of these numbers  = 490

LCM of 2x,5x,7x is 2×5×7×x = 70x

Then, we have 70x = 490

x  = 7

Hence, the numbers are 14, 35 and 49

Largest number = 49

∴The square root of the largest number = 7

5 0
2 years ago
Look at the following problem and solution given by a student: Abel, Belle, and Cindy have $408 altogether. Belle has $7 more th
sesenic [268]
A+b+c=408
a=a
b=7+a
c=5+a
a+(7+a)+(5+a)=408
3a+12=408
3a=396
a=132

Abel has $132
Belle has $139
Cindy has $137

Prove:
132+139+137=408
8 0
3 years ago
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