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3 years ago

Which expressions are equivalent to 4b? b+2(b+2b) 3b+b 2(2b)

Mathematics

2 answers:

laila [671]3 years ago

7 0

Answer:

3b+b and 2(2b)

Step-by-step explanation:

BRAINLIEST APPRECIATED!

THANKS!

Vsevolod [243]3 years ago

4 0

3b+b and 2(2b) would be it

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Please help <br><br>must show work

hjlf

Answer:

b) y=2x-3 y-int.=-3 x= 1/2 (think of rise over run)= up 2 over 1. C) IDK D) y=2/4+3 E)7/2-8

Step-by-step explanation:

<h2>when you solve in slope intercept form ex. 3x+4y=12 move 3x by subtracting to the other side. 3x-3x=0 cancel then your left with 4y=-3 (you moved to both sides)+ 12 divide the equation by 4. left with 4divided by 4 cancels left with y= -3+12 divided by 4= y=3/4+8</h2><h2 /><h2>Remember if positive subtract if negative number add to both sides.</h2>

3 0

3 years ago

X^3 - 8 = x - 2<br><br> Solve the equation for all values of x

kirill [66]

X³-8=x-2: Original equation.
x³-x-6=0: Subtract x and add 2 to both sides.
(x-2)(x²+2x+3): Factor the equation
x-2=0; x²+2x+3=0: Separate the expressions into equations
x=2: Add 2 to both sides
x²+2x+3=0: Quadratic Formula
x= 2, <u>-2+√-8</u>,<u>-2-√-8</u>
<u /> 2 2

8 0

3 years ago

Simplify your answer as much as possible.<br><br> :)

Nitella [24]

Answer:

y > 7

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

<u>1. Start with the first 11 inspected packages:</u>

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

<u>2. The 12th package</u>

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

<u>3. Finally</u>

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

6 0

4 years ago

Show all worked identify the asymptotes and state the end behavior of the function F(x)=6x over x-36

astraxan [27]

Solution

Asymptote:

Vertical Asymptote

- The vertical asymptotes of a rational function are determined by the denominator expression.

- The expression given is:

$f(x)=\frac{6x}{x-36}$

- The denominator of (x- 36) determines the asymptote line.

- The vertical asymptote defines where the rational function isundefined. Iin order for a rational function to be undefined, its denominator must be zero.

- Thus, we can say:

$\begin{gathered} x-36=0 \\ Add\text{ 36 to both sides} \\ \\ \therefore x=36 \end{gathered}$

- Thus, the vertical asymptote is

$x=36$

Horizontal Asymptote:

- The horizontal asymptote exists in two cases:

1. When the highest degree of the numerator is less han the degree of the demnominator. In this case, the horizontal asymptote is y = 0

2. When the highest degee sof the numerator and tdenominator are the same. In this case, the horizontal asymptote is

$\begin{gathered} y=\frac{N}{D} \\ where, \\ N=\text{ Coefficient of the highest degree of the numerator} \\ D=\text{ Coefficient of the highest degree of the denominator} \end{gathered}$

- For our question, we can see that the highest degrees of the numerator and denominator are the same. Thus, we have the Horizontal Asymptote to be:

$y=\frac{6}{1}=6$

End behavior:

- The end behavior is examining the y-values of the function as x tendsto negative and positive infinity.

- Thus, we have that:

$\begin{gathered} f(x)=\frac{6x}{x-36} \\ \\ \text{ Divide top and bottom by }x \\ f(x)=\frac{6x}{x-36}\times\frac{x}{x} \\ \\ f(x)=\frac{\frac{6x}{x}}{\frac{x-36}{x}}=\frac{6}{1-\frac{36}{x}} \\ \\ As\text{ }x\to-\infty \\ f(-\infty)=\frac{6}{1-\frac{36}{-\infty}}=\frac{6}{1+\frac{36}{\infty}}=\frac{6}{1+0}=6 \\ \\ \text{ Thus, we can say: }x\to-\infty,f(x)\to6 \\ \\ Also, \\ As\text{ }x\to\infty \\ f(\infty)=\frac{6}{1-\frac{36}{\infty}}=\frac{6}{1-0}=6 \\ \\ \text{ Thus, we can also say: }x\to\infty,f(x)\to6 \end{gathered}$

Final Answers

Asymptotes:

$\begin{gathered} \text{ Vertical:} \\ x=36 \\ \\ \text{ Horizontal:} \\ y=6 \end{gathered}$

End behavior:

$\begin{gathered} As\text{ }x\to-\infty,f(x)\to6 \\ \\ As\text{ }x\to\infty,f(x)\to6 \end{gathered}$

7 0

1 year ago