72/12 ghhsnjdjdjjdjbdbdbdbdbxb

Which statement about the amount of paint found by

stepan [7]

Circumference=37.68cm
2πr=37.68
6.28r=37.68
r=6cm

So

TSA of one hemisphere

2πr²+πr²
6.28(6)²+3.14(6)²
36(6.28)+36(3.14)
36(6.28+3.14)
36(9.42)
339.12cm³

For 2 hemisphere

TSA=2(339.12)=678.24cm³

3 0

2 years ago

Find the number of gallons in 6 liters of solution(1gal=3.79l)

ohaa [14]

6 divided by 3.79 = 1.583 gallons

3 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

In a movie theater, there are children and adults in the ratio 4:7. If there are 21 adults, how many people total are there in t

lozanna [386]

Answer:

33

Step-by-step explanation:

7x3=21

4x3=12

21+12=33

8 0

2 years ago

Determine the next three terms in the following sequence: 2, 3, 5, 8, ____, ____, ____,... a. 13, 19, 26 b. 12, 17, 23 c. 12, 16

Olenka [21]

Answer:

B. 12,17,23

Step-by-step explanation:

2,3,5,8, _,_,_

2,3 is increased by 1

3,5 is increased by 2

5,8 is increased by 3. theres our pattern, each time it increases it follows the next number so the next 3 numbers it would be increased by is. 4,5,6.

Now knowing it is 4,5,6 we need to add, start adding.

8+4=12. Which answers have 12 in them? B and C.

We know 12 is the next number, now we add 5 to that.

12+5=17.

B has 17 but C doesn't, so that tells you its most likely going to be B. To make sure add 6 to 17.

17+6=23. Its true that it is B, as B is 12.17.23.

4 0

2 years ago