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Fiesta28 [93]
3 years ago
13

the distant around a track is 1/4 mile. a marathon is 26 1/5 miles. How many laps does it take to run a marathon​

Mathematics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

108 1/8

Step-by-step explanation:

1/4=0.25, 1/5=0.2

=>26+0.2=26.2

=>26.2/0.25=108.8

You might be interested in
Ray has found that his new car gets 31 miles per gallon on the highwayand 26 miles per gallon in the city. He recently drove 285
Mkey [24]

Answer:

Highway = 155\ miles

City = 130\ miles

Step-by-step explanation:

Given

Highway = 31mi/gallon

City = 26mi/gallon

Total\ Miles = 285

Total\ Gallons = 10

Required

Determine the number of miles driven on the highway and on the city

Represent the gallons used on highway with h and on city with c.

So, we have:

c + h = 10 ---- gallons used

and

31h + 26c = 285 --- distance travelled

In the first equation, make c the subject

c = 10 - h

Substitute 10 - h for c in the second equation

31h + 26c = 285

31h + 26(10 - h) = 285

Open bracket

31h + 260 - 26h = 285

Collect like terms

31h - 26h = 285 - 260

5h = 25

Make h the subject

h = \frac{25}{5}

h = 5

Substitute 5 for h in c = 10 - h

c = 10 - 5

c = 5

If on the highway, he travels 31 miles per gallon, then his distance on the highway is:

Highway = 31 * 5

Highway = 155\ miles

If in the highway, he travels 26 miles per gallon, then his distance on the highway is:

City = 26 * 5

City = 130\ miles

8 0
3 years ago
In which sentence is the fraction written correctly?
klemol [59]
I think it's A.cody ate two-thirds of the apple pie
7 0
3 years ago
A Crane Lifts a load of 4000N to a height of 20m in 5 seconds. Calculate the power of the Crane in horse power​
Digiron [165]

Answer:

2.54379284 hp.

Step-by-step explanation: Torque is defined as the engines rotational force.  First we find the distance =  m in equation- then we use the formula Power = work/time = 16.7/ 5 = 3.34 and Horsepower formula is given by. Horsepower = (Torque × Speed) / 5252. HP = (3.34 × 4000) / 5252. HP = 13360/5252. HP = 2.54379284 hp.  

Given :    W=4000    m=?        h=20 m                      

Time of operation         t =5 s

Work done in lifting a box            W=mgh= ? ×12×20=4000 = m = 4000/240=16.6666667 = 16.7KG

Power required         P=  W/t   =  w/p = 120000/120 =1000 watts    We divide 60/5 = 120 then divide by 10 = 12 (being g)

3 0
2 years ago
18. The temperature was
Alex_Xolod [135]

Answer:

Step-by-step explanation: Find the number between 40 and -10.

start at 40 and end at -10.

-40=0

0-10=-10

the tempature changed by 50 degrees.

6 0
2 years ago
The area of a rectangle is 56 cm. the length is 2 less than x and the width is 5 less than twice of x. solve for x
Nostrana [21]

Answer:

x=\frac{9+\sqrt{449}}{4}

Step-by-step explanation:

The area of a rectangle is:

A=l\cdot w

First let's write both the length and the width in terms of x:

l=x-2

<u>How we obtained the answer</u>:

The question states the length is "2 less than x". In this case, "less" is a keyword that represents subtraction. Therefore, "2 less than x" is "x-2".

w=2x-5

<u>How we obtained the answer</u>:

The question states the width is "5 less than twice of x". First, start by figuring out what "twice of x" is. The keyword is "twice" which represents multiplication. Therefore, "twice of x" can be written as "2x". Then, using the same keyword from before ("less"), "5 less than..." represents "2x-5".

Given:

Area = 56 cm

l=x-2

w=2x-5

Find:

x

A=l\cdot w

56=(x-2)(2x-5)

FOIL (Firsts, outsides, insides, lasts):

56=2x^2-4x-5x+10\\56=2x^2-9x+10

Subtract 10 from both sides:

46=2x^2-9x

Subtract 46 from both sides (forms a quadratic equation):

0=2x^2-9x-46

Use the quadratic formula:

{x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}  where 0=ax^2+bx+c

Therefore:

x=\frac{9\pm\sqrt{81-4(2)(-46)}}{2(2)}\\x=\frac{9\pm\sqrt{81+368}}{4}\\\\x=\frac{9\pm\sqrt{449}}{4}

The answer is:

x=\frac{9+\sqrt{449}}{4}

This is because if we used subtraction, we would get a negative value, and we cannot have a negative length/width.

4 0
2 years ago
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