Answer:
go back to hong kong
imma send you back to hong kong
Step-by-step explanation:
Answer:
navy blue black and marron
Step-by-step explanation:
Answer:

Step-by-step explanation:
Rn(x) →0
f(x) = 10/x
a = -2
Taylor series for the function <em>f </em>at the number a is:

............ equation (1)
Now we will find the function <em>f </em> and all derivatives of the function <em>f</em> at a = -2
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = -10.2/x³ f"(-2) = -10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
f""(x) = -10.2.3.4/x⁵ f""(-2) = -10.2.3.4/(-2)⁵
∴ The Taylor series for the function <em>f</em> at a = -4 means that we substitute the value of each function into equation (1)
So, we get
Or 
B. Up
Because a positive ‘a’ value will make the parabola open upwards, and because the power is 2 (even) both sides will open up in a ‘U’ shape
Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.
Therefore, the angle BKQ is equal to 180-50-45=85°.
Of course angle BKP=180-85=95°.
Hope this helps :)