A scuba diver starts at 85 3/4 meters below the surface of the water and descends until he reaches 103 1/5

meters below sea level. ​How many meters did he descend?


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Answers

Answer 1

Answer/Step-by-step explanation:

If the Diver started at 85 3/4 below sea level (-85 3/4 from sea level.) And he descends more you have to find the difference between two distances.

In order to get to -103 1/5, how much was subtracted from the original distance of -85 3/4. To solve this use an inverse operation. Find the absolute value of the new distance and add the original distance from sea level.

I -103 1/5 I + -85 3/4 = 103 1/5 - 85 3/4 = 17.45.

The distance from each value is I 17.45 I So, the diver descended another 17.45 meters from sea level.


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x = 25!

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A(x) = 2 (3x - 2)(x - 5)

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Answer:

Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2

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