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scZoUnD [109]
4 years ago
13

You pull out the plug from the bathtub. After 40 seconds, there are 13 gallons of water left in the tub. One minute after you pu

ll the plug, there are 10 gallons left.
Assume that the number of gallons varies linearly with the time since the plug was pulled.


Write the particular equation expressing the number of gallons (g) left in the tub in terms of the number of seconds (s) since you pulled the plug.​

Mathematics
1 answer:
SCORPION-xisa [38]4 years ago
7 0

Answer:

  • g = -3/20s + 19

Step-by-step explanation:

Let's assume this is a function

<u>The points are</u>

  • (40, 13) and (60, 10)

<u>Since it is linear relation, we'll get the slope intercept form</u>

  • g = ms + b, where g- number of gallons, s- time in seconds, b- y intercept

<u>Using the points, let's calculate the formula</u>

  • m = (10 - 13)/(60 - 40) = -3/20
  • 10 = -3/20*60 + b
  • 10 = - 9 + b
  • b = 19

<u>So the formula is:</u>

  • g = -3/20s + 19
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HELP PLEASE WILL MARK BRAINLIEST!!
Arte-miy333 [17]

Answer:

Answer is option d)346.4 ft

Step-by-step explanation:

from \: the \: figure \\ AB = 300 \: ft \:  \:  \: angle \: C = 60  \:( theta)\: degress \\ in \: triangle \: ABC \: using \:  tan \: theta\\  \tan(60)  =  \frac{opposite \: side}{adjacent \: side}  \\  \tan(60)  =  \frac{300}{BC}  \\  \sqrt{3}  =  \frac{300}{BC}  \\ BC =  \frac{300}{ \sqrt{3} }  \\ BC = 100 \sqrt{3} ft \\ then \: in \: triangle \: ABD \: using \: tan \: theta \\ angle \: D = 30 \: degrees \: (theta) \\   \tan(30 )  =\frac{opposite \: side}{adjacent \: side} \\  \tan(30)   =  \frac{300}{BD}  \\  \frac{1}{ \sqrt{ 3} }  =  \frac{300}{BD}  \\ BD = 300 \sqrt{3}  \\ CD = BD - BC \\  = 300 \sqrt{3 }  - 100 \sqrt{3}  \\  = (300 - 100) \sqrt{3}  \\  = 200 \sqrt{3 }  \\  = 200 \times 1.732 \\  = 346.4 \: ft

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

7 0
3 years ago
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