If x= -4 and y=-8, which expression would have a negative value?

Mathematics

2 answers:

Brrunno [24]2 years ago

7 0

Answer:

Step-by-step explanation:

They are both negative so you will keep the negative sign when you add.

yanalaym [24]2 years ago

5 0

Step-by-step explanation:

x + y will have negative value as

x + y = -4 + (-8) = - 4 - 8 = - 12

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System of equations by substitution how do this? y= x+2 y= -4x+7

jekas [21]

Equations are given,
y= x+2

y= -4x+7

As you can see both the expressions, x+2 and -4x+7 are equal to "y". That means they are equal to each other too.

So let's equal them and solve it.

 $x+2=-4x+7\\ \\ x+2+4x=7\\ \\ x+4x+2=7\\ \\ 5x+2=7\\ \\ 5x=7-2\\ \\ 5x=5\\ \\ x=\frac { 5 }{ 5 } \\ \\ x=1$

Now we have x's value we will plug it in the first equation to find "y".

 $x=1\\ \\ y=x+2\\ \\ y=1+2\\ \\ y=3$

(1,3)

8 0

3 years ago

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.