while A is technically the same equation, when finding a value, you want all the constants to be on 1 side of the equation while the variable is on the other side

5 0

3 years ago

Suppose A and B represent two different school populations where A > B and A and B must be greater than 0. Which of the follo

Ulleksa [173]

We can reject the last one: subtracting a non-zero value will result in a smaller value.

So now we have:

<span>2(A + B)
(A + B)2
A2 + B2

If all of them are mulptiplications, then they are all equivalent!

I mean by this, if what you meant is this:

</span>
<span>2*(A + B)
(A + B)*2
A*2 + B*2

If there is no sign, then the multiplication sign is implicit,

and all of these expressions say exactly the same: two of A and two of B.
</span>

6 0

3 years ago

Match each function

Elis [28]

Answer:

* The degree of the function is 4 and the leading coefficient is positive

f(x) = (x + 6)(2x - 3)(x - 1)²

* The degree of the function is 5 and the leading coefficient is negative

f(x) = (x - 2)²(-2x - 1)²(-x + 1)

* The degree of the function is 6 and the leading coefficient is negative

f(x) = (-x + 1)³(x + 2)²(x - 3)

* The degree of the function is 5 and the leading coefficient is positive

f(x) = (-2x + 1)²(x - 3)²(x + 1)

Step-by-step explanation:

∵ f(x) = (x + 6)(2x - 3)(x - 1)²

∵ (x)(2x)(x²) = 2x^4

∴ The degree of the function is 4

∴ The leading coefficient is positive ⇒ (2)

∵ f(x) = (x - 2)²(-2x - 1)²(-x + 1)

∵ (x)²(-2x)²(-x) = (x²)(4x²)(-x) = -4x^5 ⇒ (neglect -ve with even power)

∴ The degree of the function is 5

∴ The leading coefficient is negative ⇒ (-4)

∵ f(x) = (-x + 1)³(x + 2)²(x - 3)

∵ (-x)³(x)²(x) = (-x³)(x²)(x) = -x^6

∴ The degree of the function is 6

∴ The leading coefficient is negative ⇒ (-1)

∵ f(x) = (-2x + 1)²(x - 3)²(x + 1)

∵ (-2x)²(x)²(x) = (4x²)(x²)(x) = 4x^5

∴ The degree of the function is 5

∴ The leading coefficient is positive ⇒ (4)

6 0

4 years ago