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brilliants [131]
3 years ago
13

If y’all wanna join a zoom because you bored join this

Mathematics
1 answer:
REY [17]3 years ago
5 0

Answer:

I would but i can't

Step-by-step explanation:

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WILL MARK BRAINLIEST!!!
Rus_ich [418]

Answer:

y= -2x -4

Step-by-step explanation:

slope: rise/run = 2/-1 = -2

y-intercept: -4

5 0
3 years ago
Mark and Julio are selling flower bulbs for a school fundraiser. Customers can buy bags of windflower bulbs and packages of croc
ikadub [295]

Answer:

Bag of windflower bulbs costs $8.50

Package of crocus bulbs costs $17.60

Step-by-step explanation:

Let $x be the price of one bag of windflower bulbs and $y be the price of one  package of crocus bulbs.

1. Mark sold 2 bags of windflower bulbs for $2x and 5 packages of crocus bulbs for $5y. In total he earned $(2x+5y) that is $105. So,

2x+5y=105

2. Julio sold 9 bags of windflower bulbs for $9x and 5 packages of crocus bulbs for $5y. In total he earned $(9x+5y) that is $164.50. So,

9x+5y=164.50

3. You get the system of two equations:

\left\{\begin{array}{l}2x+5y=105\\ \\9x+5y=164.50\end{array}\right.

From the first equation

5y=105-2x

Substitute it into the second equation:

9x+105-2x=164.50

7x=164.50-105

7x=59.5

x=$8.50

So,

5y=105-2·8.5

5y=105-17

5y=88

y=$17.60

5 0
3 years ago
A standard deck of playing cards contains 52 cards, equally divided among four suits (hearts, diamonds, clubs, and spades). Each
Svetradugi [14.3K]
The order does not matter, so we will be using combinations here. 52C1 is the first one since there is only one possibility. 16C1 is the second. You have to multiply these two, so 52 X 16 = 832.
3 0
3 years ago
Hey lol um pls help i don’t get it
frutty [35]
I can’t see it it’s blurry be
8 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
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