For the answer to the question above asking to f<span>ind the coordinates of Z without using any new variables.
</span>
Vector WZ equals vector VP, which is (p, -q)
So Z is (-p - r + p, q - q) which is (-r, 0)
I hope my answer helped you.
<span>We know that
there is no universal acceptance meaning of a percentile. When someone told you
that you are in the 80th percentile, the meaning of that is you
have achieved the lowest score that is greater than 80 percent of the score. It is calculated by using the formula <span>R = P/100 x (N + 1)</span></span>
Answer:
2.5
Step-by-step explanation:
1. Remove parentheses
-3x^2 + x^4 + x + 2x^4 - 7 + 4x
2. Collect like terms
-3x^2 + (x^4 + 2x^4) + (x + 4x) - 7
3. Simplify
-3x^2 + 3x^4 + 5x - 7
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
