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3 years ago

Evaluate |x|/x if x is a non zero interger

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

4 0

If $x$ is positive, then $|x|=x$ and $\dfrac{|x|}x=1$ .

If $x$ is negative, then $|x|=-x$ and $\dfrac{|x|}x=-1$ .

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PLEASE HELP!!!

Sholpan [36]

P=12T

Q=66T

-----------

Toys added to each bag:

P=12T+nT

Q=66T+nT

Therefore, 66T+nT=3(12T+nT)

------------

66T+nT=36T+3nT

T(66+n)=T(36+3n)

66+n=36+3n

3n-n=66-36

2n=30

Therefore, n=30/2=15

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*She added 15 toys to each bag.

6 0

3 years ago

How to get answer / show work

Ne4ueva [31]

48 books hope this helps

3 0

4 years ago

a triangular sign has a base that is to be 4 feet less than twice its height. A local zoning ordinance restricts the area of the

SCORPION-xisa [38]

Answer:

The base is 8 ft and the height is 6 ft.

Step-by-step explanation:

If the height of the triangle is x feet its base = 2x - 4 feet.

So the area of the triangle

= 1/2 * base * height

= 1/2 * (2x - 4) * x = 24

(x - 2) * x = 24

x^2 - 2x - 24 = 0

(x - 6)(x + 4) = 0

x = 6, -4 As x is the height it must be 6 feet.

The base = 2x - 4 = 2(6) - 4 = 8 feet.

8 0

3 years ago

Mr. Cooper wet to the market looking to purchase some fried chicken meals to bring home for

vaieri [72.5K]

Answer:

I don't believe there is enough information to answer the question.

Step-by-step explanation:

Either Mr. Cooper buys two 4 for 19$ specials for 38$ and has two left over, or he buys one special and then buys two individual fried chicken meals. The question does not tell us how much an individual chicken meal costs.

3 0

3 years ago

Read 2 more answers

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a

Masja [62]

1. $f$ has a horizontal asymptote at $y=-4$

This means that

$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

(for at least one of these limits)

2. $f$ has a vertical asymptote at $x=3$

This means that $f$ has a non-removable discontinuity at $x=3$ . Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$ -intercept at (1, 0)

This means $f(1)=0$ .

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$ . The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.