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elixir [45]
3 years ago
11

Solve the equation and show your work please 5s=7s+1-2s

Mathematics
2 answers:
cluponka [151]3 years ago
6 0
Group like terms
5s=7s+1-2s
5s+7s-2s+1
combine like terms
5s=5s+1
minus 5s from both sides
0=1
uh oh
false


no solution
Helen [10]3 years ago
4 0
Does this have certain answers?
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Find the rule and the graph of the function whose graph can be obtained by performing the translation 3 units left and 2 units d
Vsevolod [243]

Let

g(x)--------> the translation of the function f(x)

we know that

the rule of the translation is

3 units left and 2 units down

that means

(x,y)-------> (x-3,y-2)

so

In the function f(x) the point (0,0) is equal at the point  (-3,-2) in the function g(x)

therefore

the function g(x) is equal to

g(x)=f(x+3)-2

g(x)=(x+3)^{3} -2

<u>The answer is</u>

a) the rule of the translation is (x,y)-------> (x-3,y-2)

b) The graph in the attached figure


5 0
3 years ago
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saul85 [17]
There is no question
6 0
3 years ago
Read 2 more answers
Write the rule for the following arithmetic sequence: 11, 15, 19, 23, …
miv72 [106K]
11,15,19,23

an = 11 + 4(n-1)
an = 11 + 4n - 4
an = 4n + 7
5 0
3 years ago
Find the range of the data in the box plot below.
kifflom [539]

Answer:

16

Step-by-step explanation:

22-6=16 ;0

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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