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Elden [556K]
2 years ago
12

HELP PLZZZ ANSWER THIS PLZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Mathematics
2 answers:
Kruka [31]2 years ago
6 0

Answer:

C

Step-by-step explanation:

Your looking dor the distance you will reach him

Ostrovityanka [42]2 years ago
5 0

Answer:

there is no question

Step-by-step explanation:

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6(m +2 + 7m <br> Please help please help
NISA [10]

Answer:

i think that is 2m + m14

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Twenty-six out of 50 teachers drive their cars to school. Approximately what percent of these teacher drive to school
Mars2501 [29]

Answer:

52%

Step-by-step explanation:

26÷50×100

that's it, hope it's okay

4 0
3 years ago
Find 2x2 − 2z4 + y2 − x2 + z4 if x = −4, y = 3, and z = 2.
maw [93]
If you put in the substitutions it would be 2*2-(2*2*4)+(3*2)-(-4*2)+(2*4).
Simplified further by multiplying it would be 2*2-8+6--8+8
the negative 8 could be simplified further since the negatives cancel out so you'll have 2*2-8+6+8+8.
Then using the order of operations you would multiply the 2's together first to get 4 so you have 4-8+6+8+8. 
After that it's simple addition giving you an answer of -26.
I'm not sure if you're looking for the final answer or just the equation with substitutions but there's both.

3 0
3 years ago
What number must be added to 36 to obtain −12
dezoksy [38]
The correct answer is -48

36 + x = -12
X = -12 - 36
x = -48

36 + (-48) = -12
36 - 48 = -12
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&#10;

So,

&#10;f'(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0&#10;\\&#10;\\ \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&#10;\\&#10;\\ \Leftrightarrow 30x - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30(x - 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow x - 2 \ \textgreater \  0&#10;\\&#10;\\ \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 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
3 0
3 years ago
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