The best answer for this question is A.
Best of luck.
Answer:
Vijay present age: a
Gautam present age : b
2/3=a/b
3a = 2b
3/4 = (a+5)/(b+5)
4a+20 = 3b +15
(3a - 2b =0) x 3---------9a - 6b = 0
(4a - 3b = -5) x 2------- 8a - 6b = -10
9a - 6b - 8a + 6b = 10
a = 10
3(10) - 2b =0
-2b = -30
b = 15
So
Vijay present age = 10
Gautam present age = 15
Answer:520
Step-by-step explanation:.
Answer:
StartRoot 53 EndRoot units
XY = √53
Step-by-step explanation:
Choose which is point 1 and point 2 so you don't confuse the coordinates.
Point 1 (–4, 0) x₁ = –4 y₁ = 0
Point 2 (3, 2) x₂ = 3 y₂ = 2
Use the formula for the distance between two points.
Therefore the line of segment XY is √53.
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
