Need help please!!!!!!!
Need help please!!!!!!! - 1

Answers

Answer 1
Answer:

Answer:

First Option {-2 , 0 , 2}

Step-by-step explanation:

A critical point occurs when the derivative is 0 or undefined.

The function in the question is undefined at x = 2 and at x = -2 because if we put x = 2 or x = -2 the denominator becomes 0 hence anything divided by 0 is undefined, and the derivative equals 0 at x = 0 so the critical points are

{-2 , 0 , 2}

Use the quotient rule to find the derivative of the function then equal it to zero

y=(x^2-1)/(x^2-4)\n\n(dy)/(dx)=((x^2-4)(2x)-(x^2-1)(2x))/((x^2-4)^2)\n\n0=((x^2-4)(2x)-(x^2-1)(2x))/((x^2-4)^2)\n\n0=2x^3-8x-2x^3+2x\n0=-6x\nx=0

Answer 2
Answer: I believe it’s the last one
Mark brainliest if satisfied

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Every two out of three kids sang at the talent show. What percent of the kids did not sing at the talent show? (round)

Answers

33.3% do notsing in the talent show.

What is a percentage?

The percentage is calculated by dividing the required value by the total value and multiplying by 100.

Example:

Required percentage value = a

total value = b

Percentage = a/b x 100

Example:

50% = 50/100 = 1/2

25% = 25/100 = 1/4

20% = 20/100 = 1/5

10% = 10/100 = 1/10

We have,

Total number of kids = 3

Number of kids who sang = 2

The number of kids who do notsing.

= 3 - 2

= 1

The percentage of kids who do not sing.

= 1/3 x 100

= 33.33%

Thus,

33.3% do notsing in the talent show.

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66.7 %
convert to decimal the multiply by 100

AB and BA name the same ray.
Always
Never
Sometimes

Answers

Answer:

AB in terms of rays, can never be BA.

Step-by-step explanation:

AB and BA name the same ray.

No, this is never true. A ray starts from a point and move in a single direction to infinity.

So, if starting point is A and ray is moving towards B, so it will continue moving towards the points further from B.

Therefore, AB as in terms of rays, can never be BA.

They are never the same ray.  ray AB begins at point A and goes through point B to infinity and ray BA begins at point B and goes through point B to infinity

The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs at 3:30 p.m. What is a cosine function that models the variation in inches above and below the water level as a function of time in hours since 9:15 a.m.?

Answers

52 inches - 12 inches = 40 inches
amplitude:  a = 40 inches / 2 = 20 

f(x)=20cos(bx)+c
the value of c is 32... since the centre of the has been moved up 32 units

the minimum amplitude =  32 - 20 = 12
the maximum amplitude = 32 + 20 = 52

f(x)=20cos(bx)+32
if the curve takes 6 1/4 hours from low to high tides (9:15 am to 3:30 pm)  then it will take 12 1/2 hours to complete a full cycle.

adjust the period by converting 12 1/2 hours to an angle measure.

360
°/12 = 30° 
30° / 12 = 15°

12 1/2 = 360° + 15° = 375°

f(x) = 20 cos(375°) + 32
f(x) = 20 * 0.97 + 32
f(x) = 19.4 + 32
f(x) = 51.4

The cosine function that models the variation is f(x) = 51.4

Calculations and Parameters:

To find the inches, we would subtract the value of 12 from 52 which would give us 40 inches.

The amplitude is 40/2

= 20

Hence,

f(x)=20cos(bx)+c

  • c= 32
  • the minimum amplitude = 12
  • the maximum amplitude = 52

f(x)=20cos(bx)+32

if the curve takes 6 1/4 hours from low to high tides (9:15 am to 3:30 pm)  then it will take 12 1/2 hours to complete a full cycle.

We make adjustments and convert

  • 360°/12 = 30°
  • 30° / 12 = 15°
  • 12 1/2 = 360° + 15° = 375°

f(x) = 20 cos(375°) + 32

f(x) = 20 * 0.97 + 32

f(x) = 19.4 + 32

f(x) = 51.4

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Express the ratio 20:35 in simplest form.

Answers

Answer:4:7

Step-by-step explanation:

20:35

/5 on both sides.

4:7

Why might a person need an emergency plan?

Answers

that way they have something to fall back on just in case of an emergency.

A worker is hired for $80 a day on the condition that if business is slow, the worker will only receive half pay.  At the end of 20 days, the worker receives $1,320.  On how many days was business slow?

Answers

7 slow days

You know the worker worked a total of 20 days, some at $80 per day, and some at $40 per day.  After the 20 days, the worker was paid $1320.

x = days paid full time ($80)
y = days paid half time ($40)

you have two unknowns, so you'll need two equations to solve, from the problem statement, we can derive the following relationships:

($80)x + ($40)y = $1320
x + y = 20 days

x = 20 - y
80(20 - y) + 40y = 1320
1600 - 80y + 40y = 1320
1600 - 40y = 1320
1600 - 1320 = 40y
280 = 40y
y = 280/40 = 7

thus 7 days paid half time (i.e. slow days)

verify your solution is correct
y = 7
x = 20 - 7 = 13
80(13) + 40(7) = 1320 [OK]