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3 years ago

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Which statements are true about the locations of points A and B? Select all that apply. Point A is at (0, -2). 4 Point A is at (

0, -4). х Point A is at (4,0). Point B is at (-4, -3). -8 -4 4 co А -4. Point B is at (-8,-6). Point B is at (-8, -7). B -8
PLSSSS HELP

2 answers:

Igoryamba3 years ago

4 0

Point a is at (0,-4) and point b is at (-8,-7) :)

deff fn [24]3 years ago

3 0

Answer:

B and F

Step-by-step explanation:

took the test

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Find the value of the expression 4a^4 − 2b^2 + 40 when a = 2 and b = 7

Effectus [21]

Answer:6

Step-by-step explanation:

4(2)^4 − 2(7)^2 + 40

4(16)-2(7)^2+40

64-2(49)+40

64-98+40

-34+40

6

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3 years ago

Read 2 more answers

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IrinaVladis [17]

Answer:

I think it's A

Step-by-step explanation:

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3 years ago

A bacteria culture starts with 12,000 bacteria and the number doubles every 50 minutes.

Vlada [557]

Answer:

a) $y=12000(2)^{\frac{t}{50}}$

b) Approx. 27,569 bacteria

c) About 103 minutes

Step-by-step explanation:

a)

This will follow exponential modelling with form of equation shown below:

$y=Ab^{\frac{t}{n}}$

Where

A is the initial amount (here, 12000)

b is the growth factor (double, so growth factor is "2")

n is the number of minutes in which it doubles, so n = 50

Substituting, we get our formula:

$y=Ab^{\frac{t}{n}}\\y=12000(2)^{\frac{t}{50}}$

b)

To get number of bacteria after 1 hour, we have to plug in the time into "t" of the formula we wrote earlier.

Remember, t is in minutes, so

1 hour = 60 minutes

t = 60

Substituting, we get:

$y=12000(2)^{\frac{t}{50}}\\y=12000(2)^{\frac{60}{50}}\\y=12000(2)^{\frac{6}{5}}\\y=27,568.76$

The number of bacteria after 1 hour would approximate be <u>27,569 bacteria</u>

<u></u>

c)

To get TIME to go to 50,000 bacteria, we will substitute 50,000 into "y" of the equation and solve the equation using natural logarithms to get t. Shown below:

$y=12000(2)^{\frac{t}{50}}\\50,000=12,000(2)^{\frac{t}{50}}\\4.17=2^{\frac{t}{50}}\\Ln(4.17)=Ln(2^{\frac{t}{50}})\\Ln(4.17)=\frac{t}{50}*Ln(2)\\\frac{t}{50}=\frac{Ln(4.17)}{Ln(2)}\\\frac{t}{50}=2.06\\t=103$

After about 103 minutes, there will be 50,000 bacteria

4 0

3 years ago

What is 1/4 divided by 9/10 rewritten with an important fraction?

lutik1710 [3]

Step-by-step explanation:

${:}\hookrightarrow\sf \dfrac {1}{4}\div \dfrac {9}{10}$

${:}\hookrightarrow\sf \dfrac {1}{4}\times \dfrac {10}{9}$

${:}\hookrightarrow\sf \dfrac {10}{36}$

${:}\hookrightarrow\sf \dfrac {5}{18}$

8 0

3 years ago

Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

Or

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

4 years ago