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2 years ago

Solve the inequality 2x + 8 < 5x − 4.

Mathematics

1 answer:

Lana71 [14]2 years ago

7 0

The solution to this inequality is 4 < x or x > 4.

In order to solve this, follow the steps to solving an equation with one variable. The steps are below for you.

2x + 8 < 5x - 4 ------> Subtract 2x from both sides

8 < 3x - 4 -----> Add 4 to both sides

12 < 3x -----> Divide both sides by 3

4 < x

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-4x - y = 24<br> Complete the missing value in the solution to the equation.<br> ( _,8)

Zinaida [17]

Answer:

(- 8, 8 )

Step-by-step explanation:

Given ( ?, 8 ) is a solution. To find x substitute y = 8 into the equation, then

- 4x - 8 = 24 ( add 8 to both sides )

- 4x = 32 ( divide both sides by - 4 )

x = - 8

Thus solution is (- 8, 8 )

5 0

2 years ago

Im doing IXL lol ......................................

Andrew [12]

Answer:

Ok?

Step-by-step explanation:

5 0

2 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

1 year ago

The ages of the winners of a cycling tournament are approximately bell-shaped. The mean age is 27.2 years, with a standard devia

elena-14-01-66 [18.8K]

Answer:

a. z-score = -1.2

b. The age 23 years old is -1.2 standard deviations below the mean

c. The age is not unusual

Step-by-step explanation:

a. Transformation to z-score

Mathematically;

z-score = (x - μ)/σ

where σ and μ are the standard deviation and the mean respectively.

From the question, x = 23, μ = 27.2 and σ = 3.5

Let’s put these into the equation;

z-score = (23-27.2)/3.5 = -1.2

b. Interpretation of result

The interpretation is that 23 years old is -1.2 standard deviations below the mean

c. Determination

The age is not unusual. This is because any z-value below -2 or above 2 is considered unusual

-1.2 is above -2 and thus it is not unusual

7 0

2 years ago

One tank is filling at a rate of 5 gallon per 7 hour. A second tank is 8 10

seraphim [82]

Answer:

the second one

Step-by-step explanation:

8 0

2 years ago