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rjkz [21]
3 years ago
5

Rewrite without absolute value for the given conditions: y=|x−3|+|x+2|−|x−5|, if x<−2

Mathematics
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

y=-x-4 given that x is less than -2.

Step-by-step explanation:

|x-3|=x-3 when x-3 is positive, \ge 0. So solving x-3 \ge 0 by adding 3 on both sides gives x \ge 3

|x-3|=-(x-3) when x-3 is negative, \le 0. So solving x-3 \le 0 by adding 3 on both sides gives x \le 3

x is included in the last one since these values are less than 3. This means for our problem with given condition that |x-3|=-(x-3)=-x+3.

|x+2|=x+2 when x+2 is positive, \ge 0. So solving x+2 \ge 0 by subtracting 2 on both sides gives x \ge -2

|x+2|=-(x+2) when x+2 is negative, \le 0. So solving x+2 \le 0 by subtracting 2 on both sides gives x \le -2

x is included in the last one since these values are less than -2. This means for our problem with given condition that |x+2|=-(x+2)=-x-2.

|x-5|=x-5 when x-5 is positive, \ge 0. So solving x-5 \ge 0 by adding 5 on both sides gives x \ge 5

|x-5|=-(x-5) when x-5 is negative, \le 0. So solving x-5 \le 0 by adding 5 on both sides gives x \le 5

x is included in the last one since these values are less than 5. This means for our problem with given condition that |x-5|=-(x-5)=-x+5.

Lets put it altogether now.

y=|x−3|+|x+2|−|x−5|, if x<−2

y=-x+3+-x-2-(-x+5)

y=-x+3-x-2+x-5

y=-x-x+x+3-2-5

y=-x-4

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Step-by-step explanation:

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b. L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

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d. L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25} converges to s> 0.

e. L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1} converges even s> 0.

f. L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4} converges to s> 0.

g. L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4} converges to s> 0.

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Answer:

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c.)11x8(x2 for the other side)=176. 11x4x4=176.

48x2=96. 176+176+96=448

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