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SVEN [57.7K]
3 years ago
13

PLEASE HELP I will give Brainly to however gets it right. Miguel's irrigation tank for his large garden had 15.5 gallons of wate

r before he began to
fill it at a rate of 1.4 gallons per minute.
Alejandro checked his irrigation tank and found it contained 45.5 gallons, before he
began draining his tank at a rate of 0.6 gallons per minute.
After how many minutes will the two irrigation water tanks have the same amount of
water?
Mathematics
2 answers:
BARSIC [14]3 years ago
4 0

Answer:

Fractions are used when you want to express a portion of the whole. ... If the treatment plant you work at has a tank that contains 200 gallons of chemical, how ... Write down the formula and then start writing down the various information that ... What is the volume in cubic feet of a tank that is 50 feet long, 30 feet wide, and 12 ...

Missing: Miguel's ‎15.5

Step-by-step explanation:

egoroff_w [7]3 years ago
3 0

Answer:

It will take 50 minutes.

Step-by-step explanation:

Because if you are trying to get 45.5 to 15.5 you need to drain 30 gallons of water, and to do that it will take 50 minutes because 0.6*50=30.

Hope this helps!

Pls mark brainliest if it is correct

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4) The path of a satellite orbiting the earth causes it to pass directly over two
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Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
  • Satellite is approximately <u>2441.61 km</u> above the ground.

=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

  • AB = 60 km
  • AD = x
  • CD = h

Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

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