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2 years ago

Joe is going for a walk. He walks at a speed of 3 miles per hour for 7.5 miles. For how many hours does he walk?

Mathematics

1 answer:

Ksju [112]2 years ago

6 0

Answer: 2.5 hours
explanation:
7.5/3=2.5

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Which graph represents (,y)-pairs that make the equation y = 2x true?

Veseljchak [2.6K]

Answer:

Pairs (2,4), (4,8) and (5,10) makes the equation y=2x true

Step-by-step explanation:

By using the equation y=2x, eval the following points (gathered from the attached picture)

1. (2,4) which leaves us 4=2*2

2. (4,8) which leaves us 8=4*2

3. (5,10) which leaves us 10=5*2

All points from this graph satisfies the equation y=2x

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2 years ago

the brightness of the illumination of an object varies inversely as the square of the distance from the object to the source of

Free_Kalibri [48]

I=k/d^2

4=k/d^2 and 1=k/64 so if we divide the first by the second we get:

4/1=(k/d^2)/(k/64)

4=(k/d^2)*(64/k)

4=64/d^2

d^2=64/4

d^2=16

d=4 meters

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2 years ago

What is the value of the blue dot

Sergio [31]

30000000000000000000000000000000 trust me bro

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3 years ago

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Pamela buys a shirt for $36. If tax is 8%, what is the total cost of the shirt?

VashaNatasha [74]

Answer:

i think the answer is C. $38.88

hope it helps to your question!

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2 years ago

Read 2 more answers

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago