Answer:
(x1, y1) = (1, 3)
(x2, y2) = (4, 12)
Step-by-step explanation:
y= x^2 - 4
y= 5x - 8
(substitute the value for y)
x^2 - 4 = 5x - 8
(solve the equation)
x = 1
x = 4
(substitute the values)
y = 5 × 1 - 8
y = 5 × 4 - 8
(solve the equations)
y = -3
y = 12
( the possible solutions are)
x1, y1 = 1, -3
x2, y2 = 4, 12
(check the solutions)
-3 = 1^2 - 4
-3 = 5 × 1 - 8
12 = 4^2 - 4
12 = 5 × 4 - 8
(simplify)
-3 = -3
-3 = -3
12 = 12
12 = 12
(the ordered pairs are the solutions)
Exact form: X = -4/3, 2
Decimal form: x= -1.3 repeat, 2
convergent is |r|>1
so the answer would have to be less that 1.
answer is A.
There are 6 6th roots of 64 To find them, first write 64 in polar form:
64 = 64(cos0 + isin0)
<span>If z = r(cosθ + isinθ) is a 6th root of 64, then z<span>6 </span>= 64</span> By DeMoivre's Theorem, this gives us:
<span>r6[cos(6θ) + isin(6θ)] = 64[cos0 + isin0]</span>
<span>So, r6 = 64 and 6θ = 0° + k(360°)
</span> r = 2 and θ = k(360°)/6 = k(60°), k = 0, 1, 2 ,...
If z is a 6th root of 64, then z = 2[cos(k(60°)) + isin(k(60°))], where k = 0, 1, 2, ...
1st 6th root (set k = 0): 2[cos0° + isin0°] = 2
2nd 6th root (set k = 1): 2[cos60° + isin60°] = 1 + √(3) i
3rd 6th root (set k = 2): 2[cos120° + isin120°] = -1 +√(3) i
4th 6th root (set k = 3): 2[cos180° + isin180°] = -2
5th 6th root (set k = 4): 2[cos240° + isin240°] = -1 - √(3) i
<span>6th 6th root (set k = 5): 2[cos300° +isin300°] = 1 - √(3) i</span>