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4 years ago

What is the simplified expression for the expression below? 4(3x – 2) + 6x(2 – 1)

2 answers:

8 0

When simplifying expressions, you use the distributive property and combine like terms.

4(3x - 2) + 6x(2 - 1)
12x - 8 + 12x - 6x <--- Used the distributive property for both parentheses.
6x - 8 <--- Combined like terms.

So, 6x - 8 is the answer.

MrRa [10]4 years ago

7 0

In simplified form it is...

x=1.666666667

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Evaluate -|8 - 2|? -10 -6 10

jekas [21]

Answer: -6

Step-by-step explanation:

It is -6 because it is absolute value of 8-2=6 but there is a negative sign outside the absolute value sign meaning that it will be negative anyways so it is -6

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4 years ago

the sale tax in New York is 8.25%, about how much will a 65.95 pair of shoes cost, including sales tax?

Radda [10]

Answer:

$71.39

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, 8.25%=0.0825

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$65.95+$5.44=$71.39

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3 years ago

The difference between two even numbers is positive.

Otrada [13]

Answer:

Step-by-step explanation:

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3 years ago

Brett can plow out all the driveways on his street with his new four-wheel-drive truck in 2 hours. using a snow blade on a lawn

Montano1993 [528]

1/2+1/6=1/n
4/6=1/n
4n=6
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6 0

3 years ago

Read 2 more answers

Please Help! will give brainlest!

zzz [600]

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos² $\theta$ in the place of sin² $\theta$ :

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

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

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$

In conclusion,

$\displaystyle\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

7 0

2 years ago