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3 years ago

Which expression is equivalent to (2^1/2x2^3/4)^2?

1 answer:

5 0

<span>(2^1/2x2^3/4)^2
</span><span> ((2^1/2)(2^3/4))^2
</span> ((2^1/2)^2)((2^3/4)^2)
(2)(2^3/2)
(4*2^3)^(1/2)
(2*2*2^3)^(1/2)
(*2^5/2)
The answer for this case is
b. <span>sqrt 2^5</span>

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2 years ago

Use the drawing tool(s) to form the correct answer on the provided graph. Graph the solution to this system of inequalities in t

Leya [2.2K]

The solution to the system of inequalities is (-3.375, 1.75)

<h3>How to graph the inequalities?</h3>

The system of inequalities is given as:

3y > 2x+12

2x+y < -5

Next, we plot the graph of the system using a graphing tool

From the graph, both inequalities intersect at

(-3.375, 1.75)

Hence, the solution to the system of inequalities is (-3.375, 1.75)

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2 years ago

Nonlinear Systems of Equations

zheka24 [161]

The square root and cube root identities are proved.

According to the statement

we have to find that the use of the square root identity (x − y)2 = x2 − 2xy + y2

And use of cube root identity a3 + b3 = (a + b)(a2 − ab + b2).

So, For this purpose, we know that the

A. Let us assume the two conditions.

So,

2x + 3y = 6 -(1)

4x + 7y = 8 -(2)

Here we use elimination method

So, Multiply 4 with (1) and 2 with (2)

8x + 12y = 24

8x + 14y = 16

Now eliminate x from these equations

-2y = 8

here y is -4.

and the x become

2x + 3y = 6

2x -12 = 6

2x = 18

x = 9.

B. For the use of identity $(x- y)^{2} = x^{2} - 2xy + y^{2}$

Let us assume a number 26 and 28 then fill it in the condition then

$(28- 26)^{2} = 28^{2} - 2(28)(26) + 26^{2}$

Then

$(28- 26)^{2} = 784 - 1456 + 676$

$(28- 26)^{2} = 4$

And we have to prove the cube root identity then

The identity is

$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$

Then let us assume the number 8 and 9 then

$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$

$8^{3} + 9^{3} = (8 + 9)(8^{2} - 8*9 + 9^{2})$

$8^{3} + 9^{3} = (17)(64 - 72 + 81)$

$8^{3} + 9^{3} = (17)(73)$

$8^{3} + 9^{3} = (1241)$

Hence by this way we prove the square and cube root identities.

So, The square root and cube root identities are proved.

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1 year ago