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sergejj [24]
3 years ago
13

The first part of a trip driving at 40 mph was 15 miles longer and took 30 minutes longer than the second part of the trip drivi

ng at 50 mph. What I'd the distance traveled?
Mathematics
2 answers:
Natalija [7]3 years ago
6 0

Answer:

x = 65\,mi

Step-by-step explanation:

Let assume that both trip occured at constant speed. Then, the definition of speed as the ratio of travelled distance to time is used in terms of the statement:

First trip:

\frac{x + 15\,mi}{t+0.5\,h} = 40\,\frac{mi}{h}

Second trip:

\frac{x}{t} = 50\,\frac{mi}{h}

After some algebraic handling, the following linear system is created:

x - 40\cdot t =  5\\x - 50\cdot t = 0

The solution of the linear system is:

t = 0.5\,h, x = 25\,mi

The distance travelled in both trips is:

x = 65\,mi

kenny6666 [7]3 years ago
5 0

Answer:

65 miles

Step-by-step explanation:

There are two simultaneous equations . Remeber that v=d/t, 1hour = 60 minutes.

1. 40 = (d+15)/(t+0.5)

2. 50 = d/t

solving: d= 25 miles and t=0.5 hours.

The distance traveled is:

D=d+d+15

D=65 miles

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Juan spent $7.00 of the $20.00 in his wallet. Which decimal represents the fraction of the $20.00 Juan spend?
swat32

Answer:

Juan still has $13.00 left after spending $7.00 as a percentage that is 35% and as a fraction that is 35/100 or 0.35.

Step-by-step explanation:

I hope I helped in any way.

5 0
3 years ago
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Solve the following 3 × 3 system. Enter the coordinates of the solution below.
love history [14]
The system is:

i)    <span>2x – 3y – 2z = 4
ii)    </span><span>x + 3y + 2z = –7
</span>iii)   <span>–4x – 4y – 2z = 10 

the last equation can be simplified, by dividing by -2, 

thus we have:

</span>i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
iii)   2x +2y +z = -5 


The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9   

similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:


i)    2x – 3y – 2z = 4
ii)    2x + 6y + 4z = –14

thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:

3y+2z+4=-6y-4z-14
9y+6z=-18

So we get 2 equations with variables y and z:

a)   5y+3z=-9 
b)   9y+6z=-18

now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.

Let's use elimination method, multiply the equation a by -2:

a)   -10y-6z=18 
b)   9y+6z=-18
------------------------    add the equations:

-10y+9y-6z+6z=18-18
-y=0
y=0,

thus :
9y+6z=-18 
0+6z=-18
z=-3

Finally to find x, use any of the equations i, ii or iii:

<span>2x – 3y – 2z = 4 
</span>
<span>2x – 3*0 – 2(-3) = 4

2x+6=4

2x=-2

x=-1

Solution: (x, y, z) = (-1, 0, -3 ) 


Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
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I DON'T KNOW I'M JUST TRYING TO GET SOME POINTS

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