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Kamila [148]
2 years ago
13

1 Jack and Diane want to meet for breakfast. Jack walks 1.5 mph and Diane walks 1 mph. They started 5 miles apart from each othe

r. How long does it take them to meet up?
anwers : jack take 3.33 hours and diane would take 5 hours

2 How far apart were they after 1 hour?
Mathematics
1 answer:
Alex_Xolod [135]2 years ago
3 0
1.5mph (x) = 5m
x = 5:1.5 = 3.33 [hours]

1mph (x) = 5m
x = 5:1 = 5 [hours]

After one hour, with JACK’S Speed :
5m - 1.5m = 3.5m [far apart]

After one hour, with DIANE’S Speed :
5m - 1m = 4m [far apart]


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tan x = opposite/adjacent
tan 25 = x/15
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Find the distance between the following two points: (-1,6) and (-8,9). Round to the nearest TENTH.
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Answer:

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Step-by-step explanation: because

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128,147 !!!!!!!!!!!!!
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6 0
3 years ago
consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

5 0
3 years ago
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