All categories

3 years ago

Iii) Find the values of x when y = 1

1 answer:

4 0

If x = 3 and y = - 1 , find the values of the following using in identity: (9y^2 - 4x^2)(81y^4 + 36x^2y^2 + 16x^4)
Answer · 1 vote
Formula, a^3 - b^3 = (a - b)(a^2 + ab + b^2) a^3 + b^3

You might be interested in

Kristen was asked to write each of the numbers in the expression 80,000 x 25 using exponents. Her response was (8 x 10 to the po

Gnom [1K]

Kriste respone is wrong

Solution:

Given that,

Kristen was asked to write each of the numbers in the expression 80,000 x 25 using exponents

Her response was (8 x 10 to the power of 3) x 5 to the power of 2

No, the response of Kristen is wrong

Let us first understand about exponents

An exponent refers to the number of times a number is multiplied by itself

Given expression is:

$80000 \times 25$

Here, 80000 has 4 zeros

Therefore, it can be raised as 10 power 4

$80000 = 8 \times 10^4$

Thus the expression becomes:

$80000 \times 25 = 8 \times 10^4 \times 25$

Also, 25 can be written as 5 power 2

$80000 \times 25 = 8 \times 10^4 \times 5^2$

Thus given expression is written as (8 x 10 to the power 4) x 5 power to the power 2

Thus Kristen respone is wrong, because he has raised 10 to power 3

6 0

3 years ago

VOCABULARY Is 5 in the solution set of x + 3 >8 ?

Gennadij [26K]

Answer:

d. no; If x is 5, the expression on the left simplifies to 8, making the inequality false.

3 0

3 years ago

Which number is equivalent to 0.3

Strike441 [17]

3/10 that's equivalent

8 0

3 years ago

Read 2 more answers

Someone help me out please

uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\ (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4$ .

2 step (A): simplify to obtain the final radical term on one side of the equation.

$-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2$ .

3 step (F): raise both sides of the equation to the power of 2 again.

$(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2$ .

4 step (E): simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0$ .

5 step (D): use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}$ .

6 step (C): apply the zero product rule.

$x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}$ .

Additional 7 step: check these solutions, substituting into the initial equation.

3 0

3 years ago

PLEASE HELP. I don’t get it at all and I’d really appreciate it. Thank you!

White raven [17]

Answer:

was this the whole question cause it depends on what the plane is

Step-by-step explanation:

3 0

2 years ago