under the supremacy clause, can the states ignore federal laws? (its a yes or no question not a paragraph from g00gle)

How many solutions are there to the following system of equations?

ANTONII [103]

I believe the answer is A because you cannot find x nor y

5 0

4 years ago

The following rational equation has denominators that contain variables. For this equation, a. Write the value or values of the

EastWind [94]

<h3>Answer:</h3>

a) x=3 and x=-6 will make the denominator terms zero, hence x may not have those values.

b) no solution

<h3>Step-by-step explanation:</h3>

The left side can be written over a common denominator so as to make the equation be ...

((x+6) -3(x-3))/((x-3)(x+6)) = 9/((x -3)(x +6))

(-2x +15)/((x -3)(x +6)) = 9/((x -3)(x +6)) . . . . simplify

-2(x -3)/((x -3)(x +6)) = 0 . . . . . . . . . . . . . . . . subtract the right side

-2/(x+6) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . cancel factors of (x -3)

This equation has no solution. (There is no value of x that will make -2 = 0.)

_____

<em>Comment on denominator zeros</em>

After you've seen a few instances of the equation

x + c = 0

it doesn't take long for you to recognize that the solution is ...

x = -c

That is, when a denominator factor is (x+c), the corresponding value of x that makes this factor zero is x = -c. So, a denominator factor of x+c means the domain must exclude x = -c.

Knowing this, we can write the restrictions on a rational function "by inspection" without having to do anything except mental pattern matching.

_____

<em>Comment on the solution method</em>

By rewriting the rational equation to be something of the form (a/b) = 0, we have the opportunity, as here, to cancel factors that would otherwise result in extraneous solutions. Then, we don't have to be concerned with whether or not the solutions found are extraneous.

3 0

3 years ago

RCX)=2VX SCx) = x(RSX4=

aev [14]

Given the two functions:

$\begin{gathered} R(x)=2\sqrt[]{x} \\ S(x)=\sqrt[]{x} \end{gathered}$

We need to find (RoS)(4). THis is the functional composition. We take S(x) and put it into R(x) and then put "4" into that composed function. Shown below is the process:

$(RoS)(x)=2\sqrt[]{\sqrt[]{x}}$

When we plug in "4", into "x", we have:

$\begin{gathered} (RoS)(x)=2\sqrt[]{\sqrt[]{x}} \\ (RoS)(4)=2\sqrt[]{\sqrt[]{4}} \\ =2\sqrt[]{2} \end{gathered}$

3 0

1 year ago

The triangles shown are similar. Which side of triangle PQR corresponds to side LN in triangle MNL? Triangle L N P. Side L N is

Lerok [7]

Answer:

LN corresponds to RQ

Step-by-step explanation:

If we look at the diagram we find that LN corresponds to RQ, MN corresponds to QP and ML corresponds to PR .

The both triangles are similar as PQR is 2 times of MNL .

Therefore

PQR= 2MNL

PR= 2ML

28= 2(14)

QP= 2MN

20 = 2(10)

RQ =2 LN

24= 2(12)

7 0

3 years ago

Read 2 more answers

1.Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

Genrish500 [490]

If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.

8 0

3 years ago