Answer:
a) D(t) = ![\[20*\sqrt{5} * t\]](https://tex.z-dn.net/?f=%5C%5B20%2A%5Csqrt%7B5%7D%20%2A%20t%5C%5D)
b) 178.885 miles
Step-by-step explanation:
Ship A travels north at the rate of 20 mph.
Ship B travels east at the rate of 40 mph.
After t hours, Ship A is at a distance of 20t miles from the origin.
Similarly, Ship B is at a distance of 40t miles from the origin.
(a) Distance D(t) = ![\[\sqrt{(20t)^{2}+(40t)^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B%2820t%29%5E%7B2%7D%2B%2840t%29%5E%7B2%7D%7D%5C%5D)
= ![\[\sqrt{400*t^{2}+1600 * (t)^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B400%2At%5E%7B2%7D%2B1600%20%2A%20%28t%29%5E%7B2%7D%7D%5C%5D)
= ![\[\sqrt{2000*t^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B2000%2At%5E%7B2%7D%7D%5C%5D)
=
(b) Distance between the two ships when t = 4,
= ![\[20*\sqrt{5} * 4\]](https://tex.z-dn.net/?f=%5C%5B20%2A%5Csqrt%7B5%7D%20%2A%204%5C%5D)
= ![\[80*\sqrt{5} \]](https://tex.z-dn.net/?f=%5C%5B80%2A%5Csqrt%7B5%7D%20%5C%5D)
miles
= 80 * 2.236
= 178.885 miles
Area is calculated by multiplying length by width. Since we are given the width, we can test each length and see if the product matches the rectangle's area.
(2m - 3)(4m - 4) = 8m² - 20m + 12 ≠ 8m² + 4m - 12
(8m - 3)(4m - 4) = 32m² - 44m + 12 ≠ 8m² + 4m - 12
(2m + 3)(4m - 4) = 8m² + 4m - 12
(8m + 3)(4m - 4) = 32m² + 20m - 12 ≠ 8m² + 4m - 12
The length of the rectangle must be 2m + 3. Another method of solving this would be to divide the area by the width, or 8m² + 4m - 12 ÷ 4m - 4.
Answer:
x = -2
Step-by-step explanation:
x^2 - 3x - 10 when we factorize equals
(x + 2) (x - 5)
3x^2 + 5x - 2 when we factorize equals
(3x - 1) (x + 2)
x + 2 is the common term and when we put the 2 on the other side of the equation we get x = -2
El conjugado de a+bi es a-bi (<span>invierta el signo de la parte imaginaria)
</span>
<span>por lo tanto
</span><span>El conjugado de 3+4i es 3-4i,
</span><span>El conjugado de 6-i es 6+i</span>
