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3 years ago

Which expression is a factor of 4q^2r^3s + 8qrs?

Mathematics

1 answer:

devlian [24]3 years ago

4 0

Answer:

: 4qrs • (qr2 + 2)

Step-by-step explanation:

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((22q2 • r3) • s) + 8qrs

Pulling out like terms :

3.1 Pull out like factors :

4q2r3s + 8qrs = 4qrs • (qr2 + 2)

Final result :

4qrs • (qr2 + 2)

I am really soo sorry if the answer is wrong!

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3 years ago

A solid wooden cube has 4.35cm long.calculate the volume of the cube

Pani-rosa [81]

Answer:

82.31

Step-by-step explanation:

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3 years ago

Write 0.125 as if it was 2 to a power

Westkost [7]

Answer:

$2^{-3}$

Step-by-step explanation:

0.125 as a fraction is 125/1000 or 1/8

1 is the same as $2^{0}$ (anything to the power of 0 is 1)
8 is the same as 2³ (2 × 2 × 2 = 8)
$2^{0}$ ÷ 2³ = $2^{-3}$ (You subtract the powers: 0 - 3 = -3)

Hope this helps!

4 0

3 years ago

A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s

yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2} = \frac{1 \pm 9 }{-0.4}$

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

then the average speed initially is 20 km/h

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3 years ago