A food worker has safely cooled a large pot of soup to $70^{\circ} F \left(21^{\circ} C \right)$ within two hours. What temperature must the soup reach in the next four hours to be cooled properly?
a. $41^{\circ} F \left(5^{\circ} C \right)$
b. $45^{\circ} F \left(7^{\circ} C \right)$
c. $51^{\circ} F \left(11^{\circ} C \right)$
d. $55^{\circ} F \left(13^{\circ} C \right)$
Answer (3)
Using the first two values in the table, we can first find the slope to write an equation of a line, so we have
[4754 - 4962 ] / [20 - 12 ] = -26
So we have
y - 4754 = -26(x - 20)
y - 4754 = -26x + 520
y = -26x + 5274
Let's confirm that the amount after 50 minutes is correct
y = -26(50) + 5274 = 3974
So after 2 + 1/2 hrs (150 min) we have
y = -26(150) + 5274 = 1374 gallons ;
The equation in the slope-intercept form is y = − 26 x + 5274 and there will be 1374 gallons of water in the pool after 2 and a half hours.
The equation of a line that passes through two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is ( y − y 1 ) = x 2 − x 1 y 2 − y 1 ( x − x 1 ) .
Take points ( 12 , 4962 ) and ( 20 , 4754 ) .
Here, x 1 = 12 , y 1 = 4962 , x 2 = 20 , y 2 = 4754 .
The equation is :
( y − 4962 ) = 20 − 12 4754 − 4962 ( x − 12 )
( y − 4962 ) = 8 − 208 ( x − 12 )
8 ( y − 4962 ) = − 208 ( x − 12 )
8 y − 39696 = − 208 x + 2496
208 x + 8 y − 39696 − 2496 = 0
208 x + 8 y − 42192 = 0
8 y = − 208 x + 42192
y = − 8 208 x + 8 42192
y = − 26 x + 5274.
The slope intercept form is y = m x + c , where m is the slope and c is the y -intercept.
After 2 and a half hours or 150 minutes.
Put x = 150 in the equation y = − 26 x + 5274.
y = − 26 × 150 + 5274
y = − 3900 + 5274
y = 1374
So, there will be 1374 gallons of water in the pool after 2 and a half hours.
Learn more about slope-intercept form here:
https://brainly.com/question/21366542?referrer=searchResults
The equation for the volume of water in the pool is y = − 26 x + 5274 , where x is the time in minutes. After two and a half hours, there will be 1374 gallons of water remaining in the pool. This equation represents the constant rate at which the pool is being drained.
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