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3 years ago

(X-2)(3x-4) simplify

Mathematics

2 answers:

muminat3 years ago

8 0

Answer:

3x²-10x+8

Step-by-step explanation:

To simplify the expression we need to multiply each term in the first bracket with the expression in the second bracket.

x(3x-4)-2(3x-4)

3x²-4x-6x+8

Executing the operations to give a quadratic expression we get:

3x²-10x+8

The simplified expression is therefore

3x²-10x+8

Ket [755]3 years ago

7 0

Answer: $3x^2-10x+8$

Step-by-step explanation:

Given the expression $(x-2)(3x-4)$ , you can apply Distributive property to simplify it. But first, you must remember the multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-$

Then:

$(x-2)(3x-4)=(x)(3x)+(x)(-4)+(-2)(3x)+(-2)(-4)=3x^2-4x-6x+8$

The final step is to add the like terms. Therefore, you get:

$3x^2-10x+8$

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What two numbers make 68

Sergio039 [100]

Answer:67 and 1

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

When you reverse the digits in a certain two-digits number you increase its value by 36. What is the number if the sum of its di

Zanzabum

The number is 37. 73-37=36 and 7+3 = 10

6 0

3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

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

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

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

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

4 years ago

What is parallel to the line y=2x + 5

Lana71 [14]

Answer:

Well, any number that is not 5 for the y-intercept

Step-by-step explanation:

As you could tell not matter how I change the y-intercept it is always parallel to the line y=2x + 5

6 0

3 years ago

Rationalise the following: <br> (Root3-1) / (root3+1)

Thepotemich [5.8K]

Answer:

$2-\sqrt{3}$

Step-by-step explanation:

$\begin{aligned}\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1} &=\dfrac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\\\\ & =\dfrac{\sqrt{3}\sqrt{3}-\sqrt{3}-\sqrt{3}+1}{\sqrt{3}\sqrt{3}-\sqrt{3}+\sqrt{3}-1}\\\\ & =\dfrac{4-2\sqrt{3}}{2}\\\\ & = 2-\sqrt{3}\end{aligned}$

7 0

3 years ago