All categories

3 years ago

How to put -3x(4-6x) in distributive property

Mathematics

1 answer:

Rashid [163]3 years ago

3 0

Answer:

-12x+18x²

Step-by-step explanation:

Given: $-3x(4-6x)$

Distributive property of multiplication under which a number is multiplied by the sum of two number, the first number can be distributed to both of those number.

∴ $-3x(4-6x)$

solving it by using distributive property

= $(-3x\times 4)-(-3x\times 6x)$

= $-(12x) + 18x^{2}$

Opening parenthesis

= $-12x+18x^{2}$

Hence, -12x+18 $x^{2}$ is the result after using distributive property.

You might be interested in

Select all the correct answers. a group of scientists is conducting an experiment on the effects of media on children. they rand

Julli [10]

Answer:

This study uses random sampling

This study uses a control group

This study uses a repeated measures design

6 0

1 year ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago

What is the quotient of (x^3 - 3x^2 + 3x - 2) ÷ (x^2 - x + 1)?

Ugo [173]

Answer:

x-2

The choose (1)

Step-by-step explanation:

(x³-3x²+3x-2)÷(x²-x+1)

(x-2)(x²-x+1) ÷ (x²-x+1)

Delete (x²-x+1)

so = (x-2)

4 0

3 years ago

What number is 5.5% of 110

rodikova [14]

Answer: What number is 5.5% of 110

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A minus b minus a square plus b square

miskamm [114]

Answer:

what is the question about which topic

7 0

3 years ago