PLEASE HELP I WILL GIVE 50 POINTS

2 answers:

7 0

48+30 is 78, so we need to find the equation that is equal to that.
The first one is not enough, the second one is too much, and the third one is too much as well. The fourth on is the answer.

julia-pushkina [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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Solve for the unknown 3/15=9/x

Leni [432]

X = 3*15 = 45

Hope it helped.

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2 years ago

Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for

USPshnik [31]

$x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}$

$\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}$

6 0

3 years ago

Plzzzzzzz help this is due soon and I have no clue what I'm doing lol.

Umnica [9.8K]

Answer:

1) x - 4 is greater than or equal to 18

x is greater than or equal to 22

2) 12 + x is less then or equal to 17

x is less the or equal to 5

3) 23 is less than x + 16

7 is less than x

4) -6 + x is greater than or equal to -7

x is greater than or equal to -1

4 0

3 years ago

Read 2 more answers

What is t+8=15 write how u found out show ur work thanx plz hellppppp

Pachacha [2.7K]

The answer is the number that 't' must be in order for the equation
to be a true statement.

Here's how to find it:

Write t + 8 = 15

Then subtract 8 from each side: <em> t = 7</em>

8 0

3 years ago

Read 2 more answers

2. The table below shows that the distance d varies directly as the time t. Find the constant of variation and the equation whic

Artemon [7]

<h2><u>Problem Solving</u>:-</h2>

2. The table below shows that the distance d varies directly as the time t. Find the constant of variation and the equation which describes the relation.

<h2><u>Solution</u>:-</h2>

Since the distance d varies directly as the time t, then d = kt.

Using one of the pairs of values, (2, 20), from the table, substitute the values of d and t in d = kt and solve for k.

$\sf{\rightarrow{d = kt}}$