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AveGali [126]
3 years ago
13

Gallon of gas cost 4$ carols car can drive 20 miles per gallon. what is the cost to drive 30 miles

Mathematics
1 answer:
almond37 [142]3 years ago
3 0

Answer: It costs $6 to drive 30 miles.

Step-by-step explanation:

$4 per 20 miles

20/4 = 5

1$ per 5 miles

30/5 = 6

6 * 5 = 30

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Word problems<br><br>kind of urgent
seraphim [82]
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+271.42. +271.42
1109.42= 27x
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So Net Pay is better because what you actually get.
8 0
3 years ago
What is the value of the expression (−2)^4(−2)^-1?
Morgarella [4.7K]
(-2)^4=16
(-2)^-1=-1/2
-1/2x16=-8
Answer: -8
7 0
2 years ago
Will ran the diagonal distance across a square field measuring 40 yards on each side. James ran the diagonal distance across a r
slamgirl [31]

Will ran the longest

He ran for 56.57 yards, 13.56 yards longer than James.

Step-by-step explanation:

Step 1 :

Will ran the diagonal across a square field measuring 40 yards in each side.

The diagonal of a square can be obtained by the square root of the sum of the squares of its 2 sides [Because it forms the hypotenuse of a right angle triangle]

Hence when the side is 40 yards , the diagonal would be

\sqrt{40^{2} + 40^{2}}  = \sqrt{1600 + 1600}   = \sqrt{3200}  =56.57 yards

So Will ran for 56.57 yards

Step 2 :

James ran the diagonal of a rectangular field with 25 yards length and 35 yards width.

The diagonal of the rectangle can be obtained by the square root of the sum of squares of its length and width.

Hence when the length is 25 yards and width is 35 , the diagonal would be

\sqrt{35^{2} + 25^{2}}  = \sqrt{1225 + 625}} = \sqrt{1850} = 43.01 yards

So James ran for 43.01 yards

Step 3 :

Will ran for 56.57 yards and 43.01 yards.

Hence Will ran for longer distance of 56.57 yards, which is 13.56 yards more than James.

5 0
3 years ago
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
I NEED HELP I'M IN TIMER:
alekssr [168]

1/2x - 7 = 1 5/10

I'd go with c.

8 0
3 years ago
Read 2 more answers
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