All categories

3 years ago

Complete the following: (2x +3y)^2 = (2x)^2+2(2x)(3y) +(____)^2

Mathematics

1 answer:

FrozenT [24]3 years ago

3 0

Answer:

Complete function will be $(2x+3y)^2=(2x)^2+(3y)^2+2\times (2x)\times (3y)$

Step-by-step explanation:

We have given equation $(2x+3y)^2=(2x)^2+2\times (2x)\times (3y)+..........^2$

We have to find the function in place of blank space and complete the function

We know the algebraic equation $(a+b)^2=a^2+b^2+2ab$

So $(2x+3y)^2=(2x)^2+(3y)^2+2\times (2x)\times (3y)$

So in place of blank pace there will be $(3y)^2$

And complete function will be $(2x+3y)^2=(2x)^2+(3y)^2+2\times (2x)\times (3y)$

You might be interested in

1)<br> Find the value of x.

Stells [14]

x=8

Explanation:

the opposite side is 72, which means I have to make 9x end up equaling 72.

72÷9

////

72/9

9×8=72

<u>Therefore</u>

the value of x is 8.

6 0

2 years ago

3x(5x2)=(3x5)x2 which property of multiplication does the equation show

Natalija [7]

The Associative Property

8 0

3 years ago

What is the equation of the line that passes through the points (−2, 1) and (1, 10)?

yKpoI14uk [10]

The answer is
A. 3x - y= -7

3 0

3 years ago

What is the distance between (8, -3), (4,-7)?

Papessa [141]

Answer:

$4\sqrt{2}$ which is none of your choices....

Did you mean (8,-3) and (4,-7)?

Step-by-step explanation:

We need to first find the distance between the x's.

Then the distance between the y's.

The distance between the x's is 8-4=4.

The distance between the y's is -3-(-7)=4.

So the distance between two points $(x_1,y1)\text{ and }(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ .

So we already found x1-x2 and y1-y2 so now we have:

$\sqrt{(4)^2+(4)^2}$

$\sqrt{16+16}$

$\sqrt{2(16)$

$\sqrt{16}\sqrt{2}$

$4\sqrt{2}$

7 0

3 years ago

A rectangle is constructed with its base on theâ x-axis and two of its vertices on the parabola yequals=2525minusâxsquared2. wha

Nina [5.8K]

You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3

6 0

3 years ago