What is 5x-(x-2)>2x-4(x-8)

Mathematics

1 answer:

Dafna11 [192]3 years ago

3 0

Answer:

x > 5

Step-by-step explanation:

Step 1 :

Equation at the end of step 1 :

(4x + 2) - (2x - 4 • (x - 8)) > 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

6x - 30 = 6 • (x - 5)

Equation at the end of step 3 :

6 • (x - 5) > 0

Step 4 :

4.1 Divide both sides by 6

Solve Basic Inequality :

4.2 Add 5 to both sides

x > 5

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Is 0.069 less than 0.07

Nuetrik [128]

Yes, 0.069 is less than 0.07~

First, we compare the "one" place: 0 = 0.

Second, we compare the "tenth" place: 0 = 0.

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6 0

3 years ago

Round each decimal 32.62 tenths place

Tresset [83]

Answer:

32.6

Step-by-step explanation:

Look at the decimal in the hundredths place in 32.62

If the number is 0-4, round down. If the number is 5-9, round up.

In 32.62, the number in the hundredths place is 2. So, we round down to 32.6

3 0

2 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .