Irrational or rational? 6.5

Mathematics

2 answers:

vfiekz [6]3 years ago

6 0

Answer:

Rational

Step-by-step explanation:

Degger [83]3 years ago

4 0

Answer:

im pretty sure its is rational

Step-by-step explanation:

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Evaluate y = 2x + 1 when x= -1

Stolb23 [73]

Y = 2x + 1
y = 2(-1) + 1
y= -2 + 1
y= -1

5 0

2 years ago

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Factor each expression by factoring out the common binomial <br><br> 5a(y + 4) + 8(y + 4)

Tems11 [23]

Answer:

Factoring the term $5a(y + 4) + 8(y + 4)$ we get $(y+4)(5a+8)$

Step-by-step explanation:

We need to factor the term: $5a(y + 4) + 8(y + 4)$

Factoring:

$5a(y + 4) + 8(y + 4)$

Taking (y+4) common

$(y+4)(5a+8)$

It cannot be further factored.

So, Factoring the term $5a(y + 4) + 8(y + 4)$ we get $(y+4)(5a+8)$

6 0

3 years ago

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larisa86 [58]

The answer is 6m djsjdjjsskskksksksksksksmsmsm

8 0

3 years ago

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.