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2 years ago

11

If f(2) = 13 and f '(x) ≥ 3 for 2 ≤ x ≤ 4, how small can f(4) possibly be?

1 answer:

aniked [119]2 years ago

8 0

The slope f'(x) = [f(4) - f(2)]/(4-2)≥3,
so [f(4) - 13]/2 ≥3
f(4) -13 ≥ 6
f(4)≥19, so it can be as small as 19.

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defon

Answer:

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2 years ago

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If all of the people that live in your home bought one 360 pound hamburger from Mallie’s Sports Grill and ate 2/3 pound servings

Vsevolod [243]

We have 4 people in our home
so total bought 360 pounds of hamburger
each ate 2/3 pound every day
so 360-(amount eaten per day) times (number of days)=0 (0 because we want it to be gone)
number of days=x unknown
amount eaten per day=(number of people) times (amount eaten per person per day)
amount eaten per day=4 times 2/3 lb=8/3
360-8/3 times x=0
360-8/3x=0
add 8/3x to both sides
360=8/3x
multiply both sides by 3/8
1080/8=x
135=x
it would take 135 days or about 4 months or 1/3 year, assuming that my family doesn't get tired of it and that my reletives don't visit and take a whole bunch to bring home.

4 people takes 135 days

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3 years ago

2m+11=-73 what number does m=?

Leviafan [203]

Answer:

m=-42

Step-by-step explanation:

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4 0

2 years ago

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Ross and Ronnie are late for their favorite show . Ross walks around a rectangular park with dimensions of 150 ft by 215 ft. Ron

sashaice [31]

Answer:

Ronnie walked the shortest distance by (365 - 262.15 ) = 102.85 ft

Step-by-step explanation:

The image attached represent the rectangular park. In this case, the dimensions are;

AB = CD = 215 ft

AC = BD = 150 ft

Total distance walked by Ross = AB + BD = 215 + 150 = 365 ft

Distance walked by Ronnie = AD

∠ABD = right angle triangle, using pythagoras rules;

AD² = AB² + BD²

AD² = 215² + 150²

AD² = 68725

AD = √68725

AD = 262.15 ft

Distance walked by Ronnie = 262.15 ft

Ronnie walked the shortest distance by (365 - 262.15 ) = 102.85 ft

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3 years ago

If cos0=3/5 and θ is in quadrant IV, sin20=?

alexandr402 [8]

Answer:

$sen(2\theta) =\frac{24}{25}$

Step-by-step explanation:

We know that $cos(\theta) =\frac{3}{5}$ and θ is in quadrant IV

To find $sin(\theta)$ we use the following identity

$sin^2(\theta)=1-cos^2(\theta)$

$cos^2(\theta) =\frac{3^2}{5^2}$

$cos^2(\theta) =\frac{9}{25}$

Then

$sin^2(\theta)=1-\frac{9}{25}$

$sin^2(\theta)=\frac{16}{25}$

$sin(\theta)=\±\sqrt{\frac{16}{25}}$

In the IV quadrant the $sin(\theta)> 0$ then we take the positive solution

$sin(\theta)=\frac{4}{5}$

Finally to find $sin(2\theta)$ we use the following identity

$sen(2\theta) = 2 sen(\theta) cos(\theta)$

Finally

$sen(2\theta) = 2*\frac{4}{5}*\frac{3}{5}$

$sen(2\theta) =\frac{24}{25}$

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3 years ago