Answer:
It is -1.602 x 10^-19 C, and the smallest unit of charge
Step-by-step explanation:
The charge on an electron is . The charge on electron is negative. The quantization of electric charge is :
q = ne
q is net charge
n is no of electrons
e is the electronic charge which is . Hence, the correct option is (b).
(a) When f is increasing the derivative of f is positive.
f'(x) = 15x^4 - 15x^2 > 0
15x^2(x^2 - 1)> 0
x^2 - 1 > 0 (The inequality doesn't flip sign since x^2 is positive)
x^2 > 1
Then f is increasing when x < -1 and x > 1.
(b) The f is concave upward when f''(x) > 0.
f''(x) = 60x^3 - 30x > 0
30x(2x^2 - 1) > 0
x(2x^2 - 1) > 0
x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0
There are four regions here. We will check if f''(x) > 0.
x < -1/sqrt(2): f''(-1) = -30 < 0
-1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
0 < x < 1/sqrt(2): f''(0.5) = -7.5 < 0
x > 1/sqrt(2): f''(1) = 30 > 0
Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
(c) The horizontal tangents of f are at the points where f'(x) = 0
15x^2(x^2 - 1) = 0
x^2 = 1
x = -1 or x = 1
f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
f(1) = 3(1)^5 - 5(1)^3 + 2 = 0
Therefore, the tangent lines are y = 4 and y = 0.
Answer:
Step-by-step explanation:
It's 75° b/c of complementary angles.... understand? :∫
Answer: LAST OPTION.
Step-by-step explanation:
To solve this exercise is important to remember the following definitions:
- The sum is the result obtained when we solve an addition.
- The product is the result obtained when we solve a mutiplication.
- Given a multiplication , "a" and "b" are factors and "c" is the product.
In this case, you have this expression:
Notice that the there is an addition of two terms ( and ) inside the parentheses.
Outside the parentheses you can notice that the number is multiplying the sum of the terms mentioned before.
Therefore, you can conclude that the best description for this expression is:
<em>The product of a constant factor of seven and a factor with the sum of two terms.</em>