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2 years ago

9

How to solve -2y-4=4(y-1)

2 answers:

tangare [24]2 years ago

6 0

-2y-4=4(y-1)

Always remove the pararentheses as your first step.

-2y-4 = 4y - 4

Combine like terms.

-2y - 4y = 4 - 4

-6y = 0

Divide both sides by -6 to find y.

-6y/-6 = 0/-6

y = 0

Done

Naddik [55]2 years ago

5 0

Remove parentheses and then add like terms

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(x-25) (x-5) square root

Fittoniya [83]

x^2-30x+125 square root

6 0

3 years ago

The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she d

aksik [14]

Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

<h3 /><h3>Step 1</h3>

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

$f(t)=\left \{ {{0 ,\-t$

Consider the second function:

$f(t)=\frac{e^{-t/\mu}}{\mu}\\$

Where Average waiting time = μ = 2.5

The function f(t) becomes

$f(t)=0.4e^{-0.4t}$

<h3>Step 2</h3>

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

The probability that a costumer has to wait for more than x minutes is:

$\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt$

which is equal to 0.01

<h3>Step 3</h3>

Solve the equation for x

$\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01$

Take natural log on both sides

$ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53$

<h3>Results</h3>

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger

3 0

2 years ago

PLEASE HELP, BRAINLIEST FOR THE BEST ANSWER

Zolol [24]

Answer:

Use the distance formula to determine the distance between the two points.

Distance

=

√(x2−x1)^2 + (y2−y1)^2

Substitute the actual values of the points into the distance formula.

√ ( (−6) − 0)^2 +( (−3) − 4)^2

Subtract 0 from −6

√(−6)^2 + ( ( −3 ) −4 )^2

Raise −6 to the power of 2

√36 + ( ( −3 ) −4 )^2

Subtract 4 from −3

√36 + ( −7 )^2

Raise −7 to the power of 2

√ 36 + 49

Add 36 and 49

√85

5 0

2 years ago

1/5x+y=2/5<br> 1/0x+1/3y=1/2

kkurt [141]

A)
<span>|x + y = 5 </span>
<span>|2x - y = 7; </span>
<span>b) </span>
<span>|2x + y = 5 </span>
<span>|x - y = 2 </span>
<span>c) </span>
<span>|3x + y = 6 </span>
<span>|4x - 3y = -5 </span>
<span>d) </span>
<span>|1/(x - 1) = y - 3 </span>
<span>|x - y = -2 </span>
<span>e) </span>
<span>|(9x + 4y)/3 - (5x - 11)/2 = 13 - y </span>
<span>|13x - 7y = -8 </span>
<span>Answer: </span>c<span> and </span>e<span> has solution (1; 3)</span>

4 0

2 years ago

Read 2 more answers

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

3 years ago