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2 years ago

5(t+3)=-3.5 solve for t

Mathematics

2 answers:

mestny [16]2 years ago

8 0

Answer: t=−3.7

If I Helped, Please Mark Me As Brainliest, Have A Great Day :D

omeli [17]2 years ago

4 0

Answer: -3.7

Step-by-step explanation: distribUte then subtract then divide

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In 2000, Florida's total population was 15,982,378 people and 183 out of every 1000 residents were 65 years and older. Determine

dmitriy555 [2]

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2 years ago

A machine with velocity ratio of 5 is used to raise a load with an effort of 500N . If the machine is 80% efficient , determine

Yanka [14]

Answer:

Solutions given:

Velocity ratio V.R =5

effort =500N

efficiency =80%

magnitude of load=?

mechanical advantage [M.A ]

we have

efficiency =M.A/V.R*100%

80=M.A./5*100

80/100*5=M.A

M.A.=4

again

we have

M.A =load/effort

4=load/500

load=500*4

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2 years ago

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2 years ago

Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin loc

pshichka [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin.

Step-by-step explanation:

From the question we are told that

The speed from the point Sally is standing to the point P is $u = 3 \ ft /s$

The speed from the point P to the cabin is $v = 5 \ ft /s$

Let denote the distance from the bottom left corner to the P be y ft

Generally according to Pythagoras theorem the distance from Sally's first position to point P is mathematically represented as

$s = \sqrt{y^2 + 200^2}$

Generally the distance from that point P to the cabin is mathematically represented as

$d = 200 -y$

Generally the time it takes Sally to cover that distance s is mathematically represented as

$t_1 = \frac{s}{u}$

=> $t_1 = \frac{ \sqrt{y^2 + 200^2}}{3}$

Generally the time it takes Sally to cover that distance d is mathematically represented as

$t_2 = \frac{d}{v}$

=> $t_2 = \frac{ 200 - y }{5}$

So the total time taken is

$t= \frac{ \sqrt{y^2 + 200^2}}{3} + \frac{ 200 - y }{5}$

Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0

$\frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} } - \frac{1}{5} =0$

=> $\frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}$

=> $25y^2 = 9y^2 + 9 * 200^2$

=> $16y^2 = 360000$

=> $y = 150$

So the distance from point P to the cabin is

$d = 200 -150$

=> $d = 50 \ ft$

So the optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin

8 0

2 years ago