What are the different ways of the simulation indicates when an equation is balanced?

Mathematics

2 answers:

inysia [295]2 years ago

7 0

Answer:

There are three different ways that the equation is represented visually when it is balanced. First, the scale is at equilibrium when it is balanced. The balance turns yellow and a smiley face appears. Second, the graph shows equal amounts on both the reactant and product side of the equation. Third, within the individual molecule box, there should be the same number of each element on both the product and a reactant side of the equation.

Step-by-step explanation:

this is for pennfoster

yKpoI14uk [10]2 years ago

5 0

When an equation is balanced, the different ways that the simulation indicates are as follows:

- Happy Face appears

- The scales are balanced

- Pictures show same kind of atoms in reactants and products

- Pictures show same number of atoms in reactants and products

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If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.