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3 years ago

PLEASE HELP BRANNIEST

Mathematics

2 answers:

Goshia [24]3 years ago

7 0

i believe its graph D

tangare [24]3 years ago

3 0

First, you look at the choices. Since it says y is less than, B and C are wrong since the shaded part is greater.

Second, you look at the choices again. Since the equation does not say less than or equal to, the line should be dotted. Therefore, A can be ruled out.

Your answer is D. Graph D.

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The photo below shows the Royal Ontario Museum in Toronto, Canada. Are angles 2 and 4 alternate interior angles, same side inter

Sav [38]

Answer:

Alternate interior angles

8 0

3 years ago

Simplify 9y+2y-5y please answer

Brums [2.3K]

Answer:

$9y+2y-5y=6y$

Step-by-step explanation:

So we have the expression:

$9y+2y-5y$

To simplify, simply combine the like terms. Therefore:

$9y+2y-5y\\=11y-5y\\=6y$

Further notes:

To understand why we can combine like terms in the first place, we just need to use the distributive property. So we have the expression:

$9y+2y-5y$

Now, factor out a y from the three terms:

$=y(9+2-5)$

Do all the operations inside the parenthesis:

$y(9+2-5)\\=y(11-5)\\=y(6)=6y$

And we moved the y back to the front!

This is the same result as before. When combining like terms, this is what we're essentially doing but without doing the distributive property manually. I hope you understand a bit better on how and why we can combine like terms!

7 0

3 years ago

Read 2 more answers

Identify the vertex, focus, and directrix. y=1/8y^2

Brut [27]

If all the equations for the directrix are "x = " lines then this is a y^2 parabola. The actual equation is $x= \frac{1}{8}y^2$ . The standard form for a positive sideways-opening parabola is $(y-k)^2=4p(x-h)$ . We know from the equation that the vertex of the parabola is at the origin, or else the translation would be reflected within the parenthesis in the equation. Our equation has no parenthesis to indicate movement from the origin. The vertex is (0, 0). Got that out of the way. That simplifies our standard form down to $y^2=4p(x)$ . Let's take a look at our equation now. It is $x= \frac{1}{8}y^2$ . We could rewrite it and make it a closer match to the standard form if we multiply both sides by 8 to get rid of the fraction. That gives us an equation that looks like this: $y^2=8x$ . That means that 4p = 8, and p = 2. p is the distance that the focus and the directrix are from the vertex. Since this is a positive parabola, it opens up to the right. Which means, then, that the focus is to the right of the vertex, 2 units to be exact, and the directrix is 2 units to the left of the vertex. The formula for the focus is (h + p, k). Our h is 0, our k is 0 and our p is 2, so the coordinates of the focus are (2, 0). Going 2 units to the left of the origin then puts our directrix at the line x = -2. Your choice then as your answer is b.

6 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.