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3 years ago

PLEASE HELP OMG PLEASE DUE TOMORROW AND MY FINAL GRADE SO PLEASE JUST HELP.

Mathematics

1 answer:

uranmaximum [27]3 years ago

3 0

The second answer seems to be correct. Hope this helps :)

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Alex sold 330 candy bars, milk chocolate candy bars sold for $2 each and white chocolate candy bars sold for $3 each. Alex sold

Alinara [238K]

Answer:

<em>Alex sold 330 white chocolate bars</em>

Step-by-step explanation:

let the milk chocolate candy be x and the white chocolate candy be y. then

then 2x+3y=990----1)

x+y=330-----2)

from eq 2) ⇒ x=330-y---------3)

Putting eq 3 in eq 1

2(330-y) +3y=990

660-2y+3y=990

y=990-660=330-----------4)

Putting eq4 in eq1

2x+3(330)=990

2x+990=990

2x=990-990

2x=0

x=0---------------------------5)

4 0

3 years ago

What is the simplified form of x minus 2 over x squared plus x minus 6divided byx squared plus 5x plus 4 over x plus 4?

tatyana61 [14]

Answer:

$\dfrac{1}{x^2+4x+3}$

Step-by-step explanation:

You have asked for the simpilfied form of ...

$x-\dfrac{2}{x^2}+x-\dfrac{6}{x^2}+5x+\dfrac{4}{x}+4$

That would be ...

$\dfrac{7x^3+4x^2+4x-4}{x^2}$

We suspect that's not what you meant. Parentheses are required for grouping when you write math expressions in text form. They work best using math symbols instead of words. We think you mean

... ((x -2)/(x^2 +x -6))/((x^2 +5x +4)/(x +4))

_____

Maybe you want to simplify ...

$\displaystyle\frac{\left(\frac{x-2}{x^2+x-6}\right)}{\left(\frac{x^2+5x+4}{x+4}\right)}=\frac{(x-2)(x+4)}{(x-2)(x+3)(x+4)(x+1)}\\\\=\frac{1}{(x+1)(x+3)}=\frac{1}{x^2+4x+3}$

_____

<em>Comment on simplifying rational expressions</em>

Division of fractions works the same whether you're working with numbers or polynomials (or anything else). Dividing by something is the same as multiplying by its inverse (reciprocal).

(a/b)/(c/d) = (a/b)·(d/c)

I learned this as "invert and multiply". I've recently seen it referred to as "copy dot flip", meaning you copy the numerator, use a dot symbol to indicate multipication, then flip the denominator (make its reciprocal) to become what you're multiplying by.

8 0

3 years ago

Determine whether the origin is included in the shaded region and whether the shaded region is above or below the line for the g

kati45 [8]

Answer:

Option: D is correct.

Step-by-step explanation:

since we are given a inequality as:

$y$

Clearly from the graph of the following inequality we could see that the origin is included in the shaded region and the shaded area is below the line.

Also it could be seen that if we put the origin points i.e. (0,0) in the inequality than 0<2 and the condition is true and hence origin is included in the shaded area.

Hence, option D is true.