Please help with this asap!! :)

Mathematics

1 answer:

Llana [10]2 years ago

3 0

Answer:

x = 11

Step-by-step explanation:

Given :

AB = x + 5

BC = 2(x - 3)

AB = BC

x + 5 = 2(x - 3)

x + 5 = 2x - 6

5 + 6 = 2x - x

11 = x

x = 11

happy to help :)

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What is the range of the function y = x 2?

Leno4ka [110]

Answer:

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

Step-by-step explanation:

Given the function

$y=x^2$

We know that the range of a function is the set of values of the dependent variable for which a function is defined.

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

$a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)$

$f\left(x\right)\ge \:0$

Thus,

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.