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2 years ago

PLEASE HELP! I am doing Imagine Math and I need help with this question!

Mathematics

1 answer:

Vaselesa [24]2 years ago

5 0

The bottom left option is when it is reflected over the x axis, the second to last option on the left is when it is reflected of the y axis, and the middle option is when it is reflected over the line y = x

Hope this helps :)

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PLEASE HELP i will give brainliest if it is the correct answer i just really need to understand:(Select from the drop-down menus

Mnenie [13.5K]

(105^3) * (105^3).....keep the base and add the exponents

8 0

2 years ago

Read 2 more answers

Which value in the replacement set is a solution to this equation? 2n + 5 = 27 Replacement set: {11, 12, 13} Drag the solution i

Alchen [17]

11 is the answer to your question.

2×11=22

22+5=27

6 0

3 years ago

Masco Industries has 285,000 total employees. Of those employees, 85,500 telecommute. What percent of the company’s total employ

frosja888 [35]

Answer:

30% of the company’s total employees telecommute

Step-by-step explanation:

Masco Industries has 285,000 total employees. Of those employees, 85,500 telecommute.

To find the percentage of company’s total employees telecommute

We divide the number of employees telecommute by total employees

Then we multiply by 100 to get percentage

Percentage = $\frac{number \ of \ telecommute \ employess}{total \ employees} * 100$

Percentage = $\frac{85500}{285000} * 100$

Percentage = 0.3 * 100 = 30%

4 0

3 years ago

a bag contains six red marbles and seven blue marbles. You randomly pick a marble and then return it to the bag before picking a

tankabanditka [31]

Answer:

6/13 7/13 chance to get the marbles

Step-by-step explanation:

4 0

2 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago