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Mumz [18]
3 years ago
10

Charlie can type faster than any of his classmates. He took a test that showed he can type 700 words in 5 minutes (WPM). The fas

test recorded typing speed is shown in the chart below (we assume the speed is constant for the three minutes shown).
Does Charlie type faster than the record shown in the chart?

Minutes: 1 2 3
Words: 216 432 648

Choose 1 answer:
A. Yes, Charlie types faster than the record speed
B. No, Charlie types slower than the record speed
C. Charlie types at the same rate as the record speed
Mathematics
2 answers:
BartSMP [9]3 years ago
6 0

Answer: B

Step-by-step explanation:

If you divide the word by minutes for each one you get 216 which is for the record speed and if you divided how fast Charlie typed it would be 140 which mean that Charlie types slower then the record speed

jolli1 [7]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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Hope this answer is helpful


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3 years ago
Which number line represents the solution set for the inequality 3x<-9
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What is the solution for x in the equation?<br><br> 10x − 4.5 + 3x = 12x − 1.1
Aleksandr [31]

Answer:in first part of equation add 10x and 3x (like terms)

13x-4.5=12x-1.1

move all terms containing x to left side of equation

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x-4.5= -1.1

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Step-by-step explanation:

8 0
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Integral of 17/(x^3-125)
daser333 [38]

Answer:

17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

Step-by-step explanation:

∫ 17 / (x³ − 125) dx

= 17 ∫ 1 / (x³ − 125) dx

= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx

Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1

(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1

Match the coefficients and solve the system of equations.

A + B = 0

5A − 5B + C = 0

25A − 5C = 1

A = 1/75

B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

∫ 1 / (x − 5) dx = ln│x − 5│

The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

4 0
3 years ago
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