the equation that we can solve using the given system of equations is:
3x^5 - 4x^4 - 11x^3 + 2x^2 - 10x + 15 = 0
<h3>Which equation can be solved using the given system of equations?</h3>
Here we have the system of equations:
y = 3x^5 - 5x^3 + 2x^2 - 10x + 4
y = 4x^4 + 6x^3 - 11
Notice that both x and y should represent the same thing in both equations, then we could write:
3x^5 - 5x^3 + 2x^2 - 10x + 4 = y = 4x^4 + 6x^3 - 11
If we remove the middle part, we get:
3x^5 - 5x^3 + 2x^2 - 10x + 4 = 4x^4 + 6x^3 - 11
Now, this is an equation that only depends on x.
We can simplify it to get:
3x^5 - 4x^4 - 11x^3 + 2x^2 - 10x + 15 = 0
That is the equation that we can solve using the given system of equations.
If you want to learn more about systems of equations:
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Answer:
1/18
Step-by-step explanation:
Just go to calculator and write -9÷54-(-12÷54)
Write the missing numbers. 40 tens = ___ hundreds
40 * 10 = 400
so:
40 tens = 4 hundreds
We have to set up 2 different equations if we are to solve for 2 unknowns. The first equation is x = y + 4. One number (x) is (=) 4 more than another (y + 4). Since we have determined that x is larger (cuz it's 4 more than y), when we set up their difference, we are going to subtract y from x cuz x is bigger. The second equation then is
. In our first equation we said that x = y + 4, so let's sub that value in for x in the second equation:
. Expand that binomial to get
. Of course the y squared terms cancel each other out leaving us with 8y + 16 = 64. Solving for y we get that y = 6. Subbing 6 in for y in our first equation, x = 6 + 4 tells us that x = 10. Yay!
Answer:
so, it's 3.75 to 3
Step-by-step explanation:
15 quarters is 3 dollars and 3 quarters
