I need help pls sldaksdkalskdals

How would I simplify the equation 5u^6x^2+20wx/15x^4

Akimi4 [234]

Answer:

The simplified form is $5u^6x^2+\frac{4w x^{-3}}{3}$

Step-by-step explanation:

To simplify the expression given to use, we need to reduce the given expression in its simplest form. There should not be any possibility for the further cancelling out of the division terms, if any.

Now the expression that is given to us is:

$5u^6x^2+\frac{20wx}{15x^4}\\$

Here we will simplify it, as follows:

$5u^6x^2+\frac{20wx}{15x^4}\\=5u^6x^2+\frac{4w}{3x^3}\\=5u^6x^2+\frac{4w x^{-3}}{3}$

So this is the required simplified form.

3 0

3 years ago

the initial vertical velocity of a rocket shot straight up is 42 meters per second. How long does it take for the rocket to retu

Snowcat [4.5K]

It takes 4.3 seconds for the rocket to return to earth.

The equation is:
$0=-9.8t^2+v_0t+h_0$
where -9.8m/sec² is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height. We will go from the assumption that the rocket is launched from the ground, so h₀=0, and we are told that the initial velocity, v₀, is 42. This gives us:

$0=-9.8t^2+42t$

We will use the quadratic formula to solve this. The quadratic formula is:
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Plugging in our information we have:
$t=\frac{-42\pm \sqrt{42^2-4(-9.8)(0)}}{2(-9.8)}
\\
\\=\frac{-42\pm \sqrt{1764-0}}{-19.6}
\\
\\=\frac{-42\pm \sqrt{1764}}{-19.6}
\\
\\=\frac{-42\pm 42}{-19.6}=\frac{-42-42}{-19.6} \text{ or } \frac{-42+42}{-19.6}
\\
\\=\frac{-84}{-19.6}\text{ or }\frac{0}{-19.6}
\\
\\=4.3\text{ or }0$

x=0 is when the rocket is launched; x=4.3 is when the rocket lands.

8 0

4 years ago

if in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99%

Dahasolnce [82]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.

If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?

Given Information:

standard deviation = σ = 30 hours

confidence level = 99%

Margin of error = 6 hours

Required Information:

sample size = n = ?

Answer:

sample size = n ≈ 165

Step-by-step explanation:

We know that margin of error is given by

Margin of error = z*(σ/√n)

Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size

√n = z*σ/Margin of error

squaring both sides

n = (z*σ/Margin of error)²

For 99% confidence level the z-score is 2.576

n = (2.576*30/6)²

n = 164.73

since number of bulbs cannot be in fraction so rounding off yields

n ≈ 165

Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.

3 0

3 years ago

Convert to the given Unit.

Rufina [12.5K]

Answer:

24 fl oz is equal to

1.5 pints

24 as c would be 3

Step-by-step explanation:

3 0

3 years ago

Please help thank you so much

Licemer1 [7]

Answer:

$\Huge\boxed{x=15}$

Step-by-step explanation:

Hello There!

The angle labeled 6x is a right angle

If an angle is formed using the diameter of a circle then the opposite angle of that side is a right angle

So 6x = 90

isolate the variable

divide each side by 6

90/6= 15

6x/6=x

we're left with x = 15

5 0

3 years ago

Read 2 more answers