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Shtirlitz [24]
3 years ago
10

Solve for x: please help me

Mathematics
2 answers:
inna [77]3 years ago
5 0

Answer:

Option 2, x=10

Step-by-step explanation:

Angle 1 + Angle 2 = Complementary (add up to 90°)

  30°     +    6x      = 90°

=> Subtract 30 from both sides of "="

6x = 60°

=> Divide 6 on both sides of "="

x = 10 (to check your answer, you can substitute 10 in place of x and then add both angles up to make sure that it give you 90° as an answer => <u>30 + 6(10</u>) ).

Hope this helps!

solniwko [45]3 years ago
3 0

Answer:

x = 10

Step-by-step explanation:

We know that is a right angle which is 90 degrees.

we know one angle is 30 degrees.

So take 90- 30 = 60

now we solve for 6x

6*10 = 60 and 60+30=90

x = 10

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{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
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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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Answer: 79,200 feet

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