3x=78+x-2 is the answer. Good luck on your homework!
Answer:
55 =x
Step-by-step explanation:
x+15=2x-40
Subtract x from each side
x-x+15=2x-x-40
15 = x-40
Add 40 to each side
15+40 = x-40+40
55 =x
A = oz of the solution A
b = oz of the solution B
The scientist knows that solution A is 45% salt and solution B is 95% salt. She wants to obtain 80 ounces of a mixture that is 65% salt.
a + b = 80
0.45a + 0.95b = 0.65*80
by solving the above system of equations you'll find····
a = 48 oz
b = 32 oz
~I hope this helped~
If it doesn't make since, I can try to switch up the wording ^^
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
Answer:
(x-4) (x-3) (x+3)
Step-by-step explanation:
x^3 - 4x^2 - 9x + 36
Make 2 groups
x^3 - 4x^2 - 9x + 36
We can factor out x^2 from the first group and -9 from the second group
x^2 (x-4) - 9(x-4)
Now factor out (x-4)
(x-4) (x^2 -9)
This may be an answer choice but we can still factor
(x^2 -9) is the difference of squares
(x^2-9) = (x-3) (x+3)
Replacing this
(x-4) (x-3) (x+3)
