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3 years ago

Help me lolol skixisknwnd

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

5 0

Answer:

x=300

Step-by-step explanation:

First you had to find the scale factor of the image to do that you would have to divide 325 by 13 resulting in 25 then you would multiple 12 times 25 resuliting in 300.

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I need help with question 3

agasfer [191]

Answer:

see explanation

Step-by-step explanation:

To find the y-intercept let x = 0, in the equation

6x - 3y = 15 → (1)

x = 0 : - 3y = 15 ⇒ y = - 5 ⇒ (0, - 5) ← y- intercept

x + 3y = - 8 → (2)

x = 0 : 3y = - 8 ⇒ y = - $\frac{8}{3}$ ⇒ (0, - $\frac{8}{3}$ ) ← y- intercept

4 0

3 years ago

Jina is pumping water into a tank at a rate of 15 liters per minute. Since she started pumping water, 13 liters has splashed out

baherus [9]

Step-by-step explanation:

subtract-13 from -13 get zero

then, subtract 13 from 77 =64

then do divide 15 from 15

then divide 15 from 64 = 4.26 repeat

8 0

2 years ago

HELP PLEASE <br> Find the length of AB

igor_vitrenko [27]

Step-by-step explanation:

hopefully it makes sense and is visible

3 0

3 years ago

Polygon Q is a scaled copy of Polygon P using a scale factor of 1/2. Polygon 1 point

DiKsa [7]

Answer:

1/4

Step-by-step explanation:

The scale factor for area is the square of the scale factor for linear dimensions.

area scale factor = (1/2)² = 1/4

Polygon Q's area is 1/4 of the area of Polygon P.

7 0

3 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