If y is 12 when x is 5, find x when y is 30

1 answer:

3 0

We have no idea how 'x' responds to changes in 'y'.
We don't even know their units, to make a guess with.

-- If 'x' and 'y' are the ages of two people, then when y = 30, 'x' will be 23.

-- If they are the frequency and wavelength of a wave on the lake,
then when y = 30, 'x' will be 2 .

-- If 'x' is the weight of a rock in your front yard and 'y' is the number of pages
in today's newspaper, then 'y' has no effect on 'x'. No matter what happens
to 'y' . . . 'x' will always be 5 .

With the information given in the question, we just don't know.

We need either an equation, or else one more example.

You might be interested in

Are the lines 2y-10x=4 and y-2=5x parallel ??

aleksandrvk [35]

Well,

If the slope of the lines are the same, then the lines are parralel.

We need to manipulate 2y - 10x = 4 into y = mx + b form.

Add 10x to both sides
2y = 10x + 4

Divide both sides by 2
y = 5x + 2

Do the same thing with the other equation.

Add 2 to both sides
y = 5x + 2

y = 5x + 2

It appears that, not only are they parallel, but they lie on exactly the same line! If this was a System of Simultaneous Linear Equations, then there would be an infinite number of solutions!

5 0

3 years ago

Read 2 more answers

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f(x) = x2 – 2x + 2 f(x) = x3 – x

konstantin123 [22]

Answer:

$f(x)=x^{3}-3x^{2} +4x-2$

Step-by-step explanation:

we know that

The conjugate root theorem states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial

In this problem we have that

The polynomial has roots 1 and (1+i)

by the conjugate root theorem

(1-i) is also a root of the polynomial

therefore

The lowest degree of the polynomial is 3

$f(x)=a(x-1)(x-(1+i))(x-(1-i))$

Remember that

The leading coefficient is 1

a=1

$f(x)=(x-1)(x-(1+i))(x-(1-i))\\\\f(x)=(x-1)[x^{2} -(1-i)x-(1+i)x+(1-i^2)]\\\\f(x)=(x-1)[x^{2} -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^{2} -2x+2]\\\\f(x)=x^{3}-2x^{2} +2x-x^{2} +2x-2\\\\f(x)=x^{3}-3x^{2} +4x-2$