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Flauer [41]
3 years ago
8

Javier has a basket of oranges and apples. The number of oranges is 2 more than twice the number of apples in the basket. The di

fference of half the number of oranges and half the number of apples is 4.
Mathematics
1 answer:
Vesna [10]3 years ago
3 0
Hello!

Let's put this into a system of equations and have a represent the apples and o represent the oranges.

o=2a+2
0.5o-0.5a=4

Since we can already so what o equals (2a+2) we will substitute it into the second equation and solve for a.

0.5(2a+2)-0.5a=4
1a+1-0.5a=4
1.5a+1=4
1.5a=3
a=2

Now that we know the value of a will put a into the first equation to find o.

o=2(2)+2
o=4+2
o=6

There were 6 oranges and 2 apples.

I hope this helps!
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Answer:

its in number 2

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2 years ago
The function y = 8.5(1.1)x models the population in thousands of a town where x is the number of years after 2000. What will be
nikdorinn [45]

Answer:

The population will be at 65.45 in 2007.

Step-by-step explanation:

I graphed the equation below to find the answer.

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6 0
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



3 0
3 years ago
What is the answer to the problem shown?
goldenfox [79]
V(S)=S^3/6
S=6
V(6)=6^3/6=6^2=36
3 0
4 years ago
I WILL GET A ZERO SO PLEASE HELP! THANKS!!!
Setler79 [48]
M=6 is the correct answer
Plz mark brainliest
4 0
3 years ago
Read 2 more answers
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