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marta [7]
4 years ago
13

A small kite starts at 7.3 meters off the ground and girs up at 2.5 meters per second. A large kite starts at 12 .4 meters off t

he ground and goes up at 1.5 meters per second. When are the kites at the same height
Mathematics
1 answer:
tamaranim1 [39]4 years ago
5 0

Let "t" be the time in seconds at which both the kites are at the same height.

Now, for the small kite which starts at 7.3 meters off the ground, and has a speed of 2.5 meters per second, let the height at which it will be at the same height as the second kite be "h". Then, h can be written in terms of t as:

h=7.3+2.5t....Equation 1

Likewise, for the large kite which starts at 12.4 meters off the ground and goes up at 1.5 meters per second, the height at which it will be at the same height as the small kite will again be:

h=12.4+1.5t...Equation 2

Since the height is the same, Equation 1=Equation 2

7.3+2.5t=12.4+1.5t

Thus, t=5.1 seconds.

Now, at t=5.1 seconds, the height can be found by substituting t=5.1 to any one of the two equations (Equation 1 or Equation 2).

Thus, Equation 1 will give us:

7.3+2.5\times 5.1=20.05 meters.

Therefore, the kites are at the same height when the time is 5.1 seconds from start and the value of that same height is 20.05 meters.

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3 years ago
Solve the equation below <br> 9 = x/18 <br> x=<br> What is X?
dedylja [7]
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3 years ago
{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
lesya692 [45]

There are three possible outcomes that you may encounter when working with these system of equations:


  •    one solution
  •    no solution
  •    infinite solutions

We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

  1. 4x-2y+5z = 6
  2. 3x+3y+8z = 4
  3. x-5y-3z = 5

We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists



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