If you've driven at a consistent speed of 60 mph how much would you have traveled in 2 hours?

Mathematics

2 answers:

Natali5045456 [20]3 years ago

8 0

Answer:

$120$

Step-by-step explanation:

If you've driven 60 mile per hour, you'll have to find how many miles you can drive in two hours. To find this, you will have to multiply the miles by 2 hours. You can also add 60 miles with another 60 miles to find your answer.

$60*2=120$

$60+60=120$

So you've driven 120 miles per hour.

charle [14.2K]3 years ago

6 0

Answer:

120 Miles

Step-by-step explanation:

60 miles x 2 hours = 120 miles

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Read 2 more answers

Solve the system of equations: X+y+z=5 -3x-4y+4z=15 2x-y-4z=-8

atroni [7]

Answer:

(x, y, z) = (3, -2, 4)

Step-by-step explanation:

The attachment shows a solution using a graphing calculator where the first equation is solved for z, and that expression is substituted into each of the other two equations. The solution to that system is (x, y) = (3, -2). Using these values in the expression for z, we find ...

z = 5 -(3 +(-2)) = 4

The solution is (x, y, z) = (3, -2, 4).

Your graphing calculator and/or online tools can give you the reduced row echelon form of the augmented matrix representing the system:

$\left[\begin{array}{ccc|c}1&1&1&5\\-3&-4&4&15\\2&-1&-4&-8\end{array}\right]$

If you're solving by hand, you can do the substitution described above to eliminate z from one equation. You can eliminate z from another equation by adding the last two:

-3x -4y +4(5 -x -y) = 15 ⇒ -7x -8y = -5

(-3x -4y +4z) +(2x -y -4z) = (15) +(-8) ⇒ -x -5y = 7

This pair of equations can be solved for x and y in any of the usual ways. Using the second equation to substitute for x, for example, gives you ...

-7(-5y -7) -8y = -5 ⇒ 27y = -54 ⇒ y = -2

x = -5(-2) -7 = 3

z = 5 -(3) -(-2) = 4

_____

Additional comment

If all that is needed is a solution, we prefer a graphical or matrix approach. These take about the same amount of time. Both require a little additional effort to determine exact values when the solutions are not integers.