All categories

3 years ago

Factor the polynomial? 2x^2+4x^2+6x^3

Mathematics

1 answer:

tatiyna3 years ago

3 0

$2 x^{2} + 4 x^{2} + 6 x^{3}$
= $6 x^{2} + 6 x^{3}$
= $6 x^{2} (1 + x)$

You might be interested in

What is the area of the regular hexagon with a side length of 3 cm and an apothem of 2.6 cm? A. 15.6 cm2 B. 23.4 cm,2 C. 46.8 cm

yarga [219]

Regular hexagons have all equal sides and angles
=((3cmx2.6)/2) x 6
=( 7.8/2) x6
= 3.9 x 6 The answer is B. 23.4cm
= 23.4

7 0

3 years ago

Read 2 more answers

The figure consists of two quarter circles and a square.

Serggg [28]

The answer is b thank you

4 0

3 years ago

Read 2 more answers

Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

Differentiating with respect to time (t),

$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

$-6x + y\frac{dy}{dt}=0$

$y\frac{dy}{dt}=6x$

$\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Which is the required notation.

8 0

3 years ago

Solve for Y!!!!!!! 12x+16y=96

Veseljchak [2.6K]

12x + 16y = 96

Subtract 12x from both sides

16y = 96 - 12x

Divide each side by 16

y = 6 - 0.75x

8 0

3 years ago

Is 0.8-3 equal to -2.2

Marysya12 [62]

Answer:

Yes, they are equal.

Step-by-step explanation:

-3.0 + 0.8 = -2.2

Hope it helps.

8 0

2 years ago