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2 years ago

Help with a pre-calc question? I'll mark the Brainliest if you explain your answer.

Mathematics

1 answer:

anzhelika [568]2 years ago

3 0

Answer:

It has none. Angle A is obtuse, so it would have to be opposite the longest side. If side a is opposite angle A, then that is not the case.

Step-by-step explanation:

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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

2 years ago

(sin (x+y)) / sinxcosy = 1+ cot(x) + tan(y)

ivann1987 [24]

Answer:

sin (×+y)/sinx cosy =1+ cotx tany

from trig ratio Sin [x +y] = sinx Cosy + cosx siny

Sinx cosy + cosox siny./Sinx cosy

then check the attachments for further information

5 0

1 year ago

Liam has $250. Then, he and his classmates bought a present for their teacher, evenly splitting the $p cost among the 24 of them

Semenov [28]

Answer: The required expression will be

$\frac{6000-p}{24}$

Step-by-step explanation:

Since we have given that

Amount that Liam has = $250

Amount that he and his classmates spent on present for their teacher =$p

Number of students in which $p will split = 24

According to question,

Amount that left to Liam is given by

$250-\frac{p}{24}\\\\=\frac{6000-p}{24}$

Hence, the required expression will be

$\frac{6000-p}{24}$

4 0

2 years ago

Read 2 more answers

Solve the equation for x, where x is a real number (5 points): <br> 2x^2 + 15x + 4 = 0

vitfil [10]

To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                           x = [ -b ± √(b^2 - 4ac) ] / (2a)
                           x = [ -15 ± √((15)^2 - 4(2)(4)) ] / ( 2(2) )
                           x = [-15 ± √(225 - (32) ) ] / ( 4 )
                           x = [-15 ± √(193) ] / ( 4)
                           x = [-15 ± sqrt(193) ] / ( 4 )
                           x = -15/4 ± sqrt(193)/4
The answers are -15/4 + sqrt(193)/4 and -15/4 - sqrt(193)/4.

6 0

2 years ago

2. Given: -2x =4y +62(2y+3) =3x -5What is the solution (x,y)?I

Andrews [41]

this is a 2x2 system of equations.

Let:

-2x = 4y + 6 (1)

and

2(2y+3) = 3x - 5 (2)

Let's rewrite (1) as:

2x + 4y = -6 (1)

and (2) as:

3x - 4y = 11 (2)

Now, from (1)

Let's solve for x:

2x + 4y = -6

Subtract 4y from both sides:

2x + 4y - 4y = -6 - 4y

2x = -6 - 4y

Divide both sides by 2:

2x/2 = (-6 - 4y)/2

x = -3 - 2y (3)

Replacing (3) into (2)

3x - 4y = 11

3(-3 - 2y) - 4y = 11

Using distributive property:

-9 - 6y - 4y = 11

Add like terms:

-10y - 9 = 11

Add 9 to both sides:

-10y - 9 + 9 = 11 + 9

-10y = 20

Divide both sides by -10:

(-10y)/-10 = 20/-10

y = -2

Finally, replace the value of y into (3)

x = -3 - 2y

x = -3 - 2(-2)

x = -3 + 4

x = 1

Therefore the solution is :

x= 1 and y=-2

(x,y) = (1,-2)

7 0

8 months ago