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2 years ago

The Celsius temperature, C, of a chemical reaction increases and then decreases over time according to the

Mathematics

1 answer:

mixas84 [53]2 years ago

5 0

Answer:

See explanations below

Step-by-step explanation:

Given the quadratic formula that expresses the temperature of a chemical reaction

C(t)=-12t² +8t +93

Given that C(t) = 110°, we are to find the value of t

Substitute

110 = -12t² +8t +93

-12t² +8t +93-110 = 0

-12t² +8t -17 = 0

12t² -8t + 17 = 0

Using the general formula;

t = -(-8)±√(-8)²-4(12)(17)/2(12)

t = 8±√64-816/24

t = 8±√-752/24

<em>Since the value of t will give a complex number, hence the time doesn't exist when the reaction reaches 110°C</em>

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Vera_Pavlovna [14]

Answer:

A) T = -1.

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

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Step-by-step explanation:

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3 years ago

(47 Points)

Semenov [28]

case 1)

Line 1

Let A (2,5) B (-3,-5)

Line 2

Let C (3,0) D (0,-3)

Find the equation of the line 1

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with m=2 and the point A (2,5) find the equation of the line 1

y-y1=m*(x-x1)------->y-5=2*(x-2)------>y=2x-4+5------> y=2x+1

Find the equation of the line 2

Line 2

Let C (3,0) D (0,-3)

slope m=(y2-y1) /(x2-x1)------> m=(-3-0) /(0-3)-----> m=1

with m=1 and the point C (3,0) find the equation of the line 2

y-y1=m*(x-x1)------->y-0=1*(x-3)------>y=x-3

using a graph tool

see the attached figure N 1

the solution is the point (-4,-7)

case 2)

Line 1

Let A (1,1) B (2,3)

Line 2

Let C (0,3) D (2,5)

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Find the equation of the line 2

Line 2

Let C (0,3) D (2,5)

slope m=(y2-y1) /(x2-x1)------> m=(5-3) /(2-0)-----> m=1

with m=1 and the point C (0,3) find the equation of the line 2

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using a graph tool

see the attached figure N 2

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Line 1

Let A (1,0) B (0,-1)

Line 2

Let C (0,3) D (-2,-1)

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y-y1=m*(x-x1)------->y-3=2*(x-0)------>y=2x+3

using a graph tool

see the attached figure N 3

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Line 1

Let A (2,0) B (0,-2)

Line 2

Let C (4,5) D (3,3)

Find the equation of the line 1

Line 1

Let A (2,0) B (0,-2)

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