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2 years ago

What is the geometric mean of 4/5 and 45?

Mathematics

1 answer:

AfilCa [17]2 years ago

3 0

Answer:

Step-by-step explanation:

The geometric mean of 2 numbers a and b is $\sqrt{ab}$ , then

geometric mean = $\sqrt{\frac{4}{5}(45) }$ = $\sqrt{36}$ = 6

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A soccer team is having a car wash. The team spent $45 on supplies. They earned $350, including tips. The team's profit is the a

umka21 [38]

The profit is going to be the total amount made, so you are going to subtract $350 - $45 = $305. $305. I hope this helps!

5 0

3 years ago

Which of the following is a factor of 2x4 + 22x3 + 60x2?

alex41 [277]

For the answer to the question above,
2x^4 + 22x^3 + 60x^2

<span>GCF: 2x^2 </span>

<span>2x^2(x^2 + 11x + 30) =
So the following factor for the problem above is
2x^2(x + 6)(x + 5)
</span>
If you want to double check, just multiply the answer.
I hope my answer helped you.

3 0

3 years ago

Read 2 more answers

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

6 0

3 years ago

After the fraction x plus 1 all over 2 minus the fraction x plus 2 all over 3 x have been combined using the least common denomi

RoseWind [281]

If I've read this correctly, it looks like this.

$\dfrac{(x + 1)}{2 - \dfrac{(x + 2)}{3x}}$

If that is correct, then the first step is to put the top part of the denominator over 3x

$\dfrac{(x + 1)}{\dfrac{6x - (x + 2)}{3x}} = \dfrac{(x + 1)}{\dfrac{5x -2}{3x}}$

The next part is to flip a three tier fraction. I'm afraid I have to show what happens. My latex is not that strong.

What you get is

$\dfrac{3x*(x + 1)}{(5x - 2)}$

This is just about your final answer. You could write it as

$\dfrac{3x^2 + 3x}{(5x - 2)}$

6 0

3 years ago

On the left bottom, im not sure what to do there

egoroff_w [7]

I believe that it wants you to classify the kind of triangle each triangle is, but I’m not sure.

7 0

3 years ago