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3 years ago

Please help me :)), do not 'troll' with answers, it's graded...

Mathematics

1 answer:

wolverine [178]3 years ago

4 0

The first one, decreasing then increasing

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Will give Brainlies!!! What is the y=mx+b of the equation 3x−2y=12

andreyandreev [35.5K]

Answer:

y=3/2x-6

Step-by-step explanation:

To find the y=mx+b you need to move the y on one side while moving the others on the other side

First, move the 3x to the other side by subtracting by 3x on both sides:

-2y=-3x+12

Divide by -2 on both sides:

y=3/2x-6

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3 years ago

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Write a function rule for the situation: the length l(w) of a box that is two more than four times the width w.

dlinn [17]

The length is going to equal four times the width plus two, as the situation states.

L(w) = 4w + 2

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3 years ago

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X = 16 is a solution for the equation 17 + x = 33 true or false

lilavasa [31]

$\begin{gathered} 17+x=33 \\ 17+x-17=33-17 \\ x=16 \end{gathered}$

its true

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8 months ago

What is the simplest form of (x-2)(2x+3)

Vera_Pavlovna [14]

2x^2-4x-12

hope that helps :)

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2 years ago

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Find the center, vertices, and foci of the ellipse with equation 4x^2 + 9y^2 =36

Anvisha [2.4K]

Answer:

The center, vertices and foci of the ellipse with equation $4 x^{2}+9 y^{2}=36 is (0,0),(\pm 3,0),(\pm \sqrt{5}, 0)$ respectively

Solution:

The equation of ellipse with centre (0, 0) in the form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ --- eqn 1

Where,

x is the major axis

Centre (0, 0)

Vertices is $(\pm \mathrm{a}, 0)$

Foci is $(\pm \mathrm{c}, 0)$ where $c=\sqrt{a^{2}-b^{2}}$

Now given that the equation of ellipse is

$4 x^{2}+9 y^{2}=36$ --- eqn 2

On dividing equation (2) by 36,

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

On comparing equations (1) and (2),

We get a = 3, b= 2

$c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$

So centre of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (0, 0)$

Vertices of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm 3,0)$

Foci of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm \sqrt{5}, 0)$

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3 years ago