Answer: I say B its a bit hard but i think i got it not sure but Good Luck!
Step-by-step explanation:
Complete Question
If minor arc AB measures 9 inches, what is the length of the radius of circle C? Where the radians is 0.75 radians
If necessary, round your answer to the nearest inch. 6 inches 12 inches 18 inches 24 inches
Answer:
12 inches
Step-by-step explanation:
The formula for Arc length =
Arc length = r θ
θ = 0.75
Arc length = 9 inches
Hence:
r = 9/ θ
r = 9/0.75
r = 12 inches
Answer:
f(x)= 3x-4
f(-1)= 3*-1-4
f(-1)= 2-4
f(-1)= -2
f(x)= 3x-4
f(3)= 3*3-4
f(3)= 9-4
f(3)=5
f(x)= 3x-4
f(1)= 3*1-4
f(1)= 3-4
f(1)= -1
Step-by-step explanation:
It should be bigger, think of a pie:
if you cut it into 2 parts, the pieces are big
if you cut it into 4 parts, the pieces are smaller
in both parts, the AMOUNT OF PIE stays the same, the size changes
here, we cut out 8 / 10 pieces, (manny)
ana has the same WHOLE amount, but in BIGGER pieces because she has an amount less than 8.
It works if you REDUCE THE FRACTION, 8/10 can be REDUCED to 4/5, Ana's pieces should be twice as big as Manny's, but she only has 4
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
