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3 years ago

5x + 12 + 3x = 6(x + 4) Please Show Work.

Mathematics

2 answers:

Len [333]3 years ago

8 0

5x + 12 + 3x = 6(x + 4)

5x+3x+12=6(x+4)

Multiply the bracket by 6

(6)(x)(6)(+4)=6x+24

5x+3x+12=6x+24

8x+12=6x+24

Move 6x to the other side. Sign changes from +6x to -6x.

8x-6x+12=6x-6x+24

8x-6x+12=24

2x+12=24

Move 12 to the other side. Sign changes from +12 to -12.

2x+12-12=24-12

2x=12

Divide by 2 for both sides.

2x/2=12/2

x=6

Answer: x=6

castortr0y [4]3 years ago

5 0

The answer is x= 6
This is how I got the answer

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2 years ago

The exponential model Upper A equals 682.2 e Superscript 0.013 t describes the population, A, of a country in millions, t year

Triss [41]

Answer:

The population will become 840 million in 2019.

Step-by-step explanation:

Given:

The exponential model of population is given as:

$A=682.2e^{0.013t}$

Here, 't' is in years measured since 2003.

This means for the year 2003, t = 0 and so on.

Now, in order to get the year when the population is 840 million, we need to plug in 840 for 'A' and solve for 't'. Therefore,

$840=682.2e^{0.013t}\\\\e^{0.013t}=\frac{840}{682.2}\\\\e^{0.013t}=1.2313$

Taking natural log on both sides, we get:

$0.013t=\ln(1.2313)\\\\0.013t=0.2081\\\\t=\frac{0.2081}{0.013}\\\\t=16\ years$

Therefore, 16 years after 2003, the population will be 840 million.

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3 years ago

(5x - 4) - (10x - 3)

nikdorinn [45]

Answer:

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2 years ago

Int(1 \(1 + {e}^{x} )

Andreyy89

Answer:

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}$ .

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}$ .

In other words,

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}$ .

Apply $u$ -substitution on the integral $\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}$ :

Let $u = 1 + e^{x}$ . $u > 1$ .

$du = e^{x}\cdot dx$ .

$\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C$ .

Therefore

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}$ .

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2 years ago