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In-s [12.5K]
3 years ago
10

I REALLY NEED HELP WITH MY CALC CORRECTIONS. THIS IS THE THIRD TIME IVE POSTED THIS. I NEED HELP ASAP PLS!! I HATE CALC 1

Mathematics
1 answer:
Alex73 [517]3 years ago
3 0
1. As x approaches 1,  f(x) approaches  3-1,  that is 2.

If a = 2 and b = 3,   f(1) = 2(1)^2 + 3(1)  = 5

So there is a 'jump' in values of x at x = 1.  So its not continuous at x=1.

2.  For continuity  at x = 1, ax^2  + bx must = 2  that is when a + b = 2.

3. for continuity at x = 2 , ax^2 + bx  must be = 0 that is when a*2^2 + 2b = 0

- that is 4a + 2b = 0.

4. we have system of equations:-

a + b = 2
4a + 2b = 0 

this gives a = -2 and b = 4   So  f(x) is continuous when a = -2 and b = 4.
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3 years ago
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Manuela solved the equation 3−2|0.5x+1.5|=2 for one solution. Her work is shown below. 3−2|0.5x+1.5|=2 −2|0.5x+1.5|=−1 |0.5x+1.5
adoni [48]

Answer:

x = - 4

Step-by-step explanation:

Given

3 - 2 |0.5x + 1.5 | = 2 ( subtract 3 from both sides )

- 2 |0.5x + 1.5 | = - 1 ( divide both sides by - 2 )

|0.5x + 1.5 | = 0.5

The absolute value function always gives a positive value, however, the expression inside can be positive or negative, thus

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0.5x = - 1 ( divide both sides by 0.5 )

x = - 2 ← solution Manuela obtained

OR

-(0.5x + 1.5) = 0.5 ← distribute parenthesis on left side by - 1

- 0.5x - 1.5 = 0.5 ( add 1.5 to both sides )

- 0.5x = 2 ( divide both sides by - 0.5 )

x = - 4 ← other solution to the equation

5 0
3 years ago
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