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Reptile [31]
3 years ago
13

Which estimate best describes the area under the curve in square units?

Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0

Answer:

The answer is 20 units².


harina [27]3 years ago
6 0

Answer:

Step-by-step explanation:

To determine the best estimate for the area under the curve we can use the area of boxes concept. It means we can estimate it by adding the area of the boxes .

It is easier to find the area of boxes as they are forming a rectangle .

Notice that

the length of the shaded area is 4 units (from 0.5 to 4.5)

Similarly the width of the shaded area is roughly 5 units (0 to 5)

we know that the area of a rectangle is length times width .

Rectangle because the boxes are roughly forming a rectangle with  l = 4 and w = 5

so area = 4·5

Area = 20 units ^2

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What is the length in centimeters of EF?
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An investor has an account with stock from two different companies. Last year, his stock in Company A was worth $6830 and his st
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Step-by-step explanation:

An investor has an account with stock from two different companies. Last year, his stock in Company A was worth $6830 and his stock in Company B was worth $2600. The stock in Company A has increased 20% since last year and the stock in Company B has increased 5%. What was the total percentage increase in the investor's stock account? Round your answer to the nearest tenth (if necessary).

5 0
2 years ago
A(x) = 2 (3x - 2)(x - 5)
PSYCHO15rus [73]

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

Step-by-step explanation:

7 0
2 years ago
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