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3 years ago

Simplify ( √ 5)(^3 √ 5) the answer will be a wholenumber and a fraction

Mathematics

1 answer:

arlik [135]3 years ago

6 0

Answer:

$\sqrt{5}\cdot\sqrt[3]{5} =\sqrt[6]{5^3} \cdot\sqrt[6]{5^2} =\sqrt[6]{5^5} =5^{(5/6)}$

Step-by-step explanation:

The rules of exponents apply, even when they are fractional exponents:

$a^b\cdot a^c=a^{b+c}\\\\\sqrt[b]{x^a}=x^{(a/b)}$

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Divide £50 in the ratio of 1:4

NikAS [45]

Answer=10:40

Explanation:

Don’t know how to type the symbol in font of the 50 so I will substitute
£50=$50

So 1:4
Add it up
1+4=5

So 5 total units
$50=5
Divided by 5
$10=1 units

So 1:4=10:10 times 4=10:40

6 0

3 years ago

Is my answer correct?

just olya [345]

Yes you are correct 20% of 150 is 30.

3 0

3 years ago

Read 2 more answers

Pedro and Carmela are cleaning their kitchen. It takes Pedro 3 hours to clean the kitchen by himself, while Carmela can clean th

Igoryamba

Answer:

Step-by-step explanation:

Here, we want to choose the equation representing the time it will take if they worked together

Let the amount of tasks to be done be t

pedro will work at a rate of t/3

Camela will take t/2

So the joint rate will be

t/2 + t/3

This joint rate will be equal to 1

Thus, we have

t/2 + t/3 = 1

5 0

3 years ago

A projectile is fired from a cliff 190 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of

jenyasd209 [6]

Answer:

Step-by-step explanation:

I'm not sure if this problem is physics based or calculus based. So I used calculus because it just makes more sense to do so.

If the projectile is fired from 190 feet in the air, then h0 in our quadratic will be 190. Since we are being asked to find both the displacement in the x-dimension and in the y-dimension, we need a bit of physics here as well. The velocity in the y-dimension is found in

$v_{0y}=v_{0}sin\theta$ and the velocity in the x-dimension is found in

$v_{0x}=v_{0}cos\theta$

Since the sin and the cos of 45 is the same, it's made a bit simpler for us. The velocity in both the x and the y dimension is 35.35533906 feet per second.

We can use that now to write the quadratic we need to start solving these rather tedious problems.

The position function for the projectile is

$s(t)=-16t^2+35.35533906t+190$

I'm going to kind of mix things up a bit, because in order to find the distance in the horizontal dimension that the object is when it's at its max height, we need to first find out how long it takes to get to its max height. We will first take the derivative of the position function to get the velocity function of the projectile. The first derivative of the position function is

$v(t)=-32t+35.35533906$

Remember that the first derivative is the velocity function of the projectile. You should know from either physics or calculus that at its max height, the velocity of an object is 0 (because it has to stop in the air in order to turn around and come back down). Setting the velocity function equal to 0 and solving for time will give us the time that the object is at the max height.

$0=-32t+35.35533906$

I'm going to factor out the -32 to make things easier:

$0=-32(t-1.104854346)$ which gives us that at approximately 1.10485 seconds the object is at its max height.

Moving over to the horizontal distance question now. The displacement the object experiences in the horizontal dimension is found in d = rt. We know the horizontal velocity and now we know how long it takes to get to its max height, so the horizontal distance is found in

d = (35.35533906(1.10485) so

d = 39.06 feet When the object is at its max height the object is a horizontal distance of 39.06 feet from the face of the cliff. That's a.

Now to find the max height, we will use again how long it took to get to the max height and sub it in for t in the position function.

s(1.10485) = -16(1.10485)^2 + 35.35533906(1.10485) + 190 to get that the max height is

209 feet. That's b.

Now for c. We are asked when the object will hit the water. We know that when the object hits the water it is no longer in the air and has a height of 0 above the water. Sub in a 0 for s(t) in the original position function and factor to solve for t:

$0=-16t^2+$ 35.35533906t + 190 and solve for t by factoring however you find to be the easiest. Quadratic formula works great!

We find that the times are -2.51394 seconds and 4.72365 seconds. Since time will NEVER be negative, we know that the time it takes to hit the water is 4.72365 seconds.

Whew!!!

8 0

3 years ago

What is the argument of -5\sqrt{3}+ 5i?

Shtirlitz [24]

Answer:

150 degrees

Step-by-step explanation:

Graphing the complex number we see the angle terminates in the second quadrant. This means the argument, the angle, will be between 90 degrees and 180 degrees.

So if we create a right triangle with that point after graphing it. We see the height of that triangle is 5 because that is the imaginary part. The base of that triangle has length $5\sqrt{3}$ . The problem is this doesn't give us any part of the angle we want, but it does give us the complementary of the part of the angle that is in second quadrant.