Solve for x. 100° (2x+3) 51°

The first term of a sequence is -12. The recursive formula for the sequence is an = an-1 + 9. What are the next 3 terms in the s

Bess [88]

The next three terms are: -3, 6, and 15

To get each new term, we simply add 9 to the last term. So we add 9 to -12 to get -3. Then we add 9 to -3 to get 6. Finally we add 9 to 6 to get 15

In other words,
first term = -12
second term = first term+9 = -12+9 = -3
third term = second term + 9 = -3+9 = 6
fourth term = third term + 9 = 6+9 = 15

This process repeats forever though we only need three terms in this case.

So that's why the answer is the three values -3, 6 and 15

8 0

3 years ago

Becky says that when she simplifies the following expression, the value is 6 less than her age. What is Becky's age? (7+11) - 8÷

Hitman42 [59]

(7+11) is 18, (8/2) is 4, 18-4 is 14! If this expression is 6 less than her age, she would be 20!

6 0

2 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$