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4 years ago

Find the product. 1.5m6(-2m2)4

Mathematics

1 answer:

kogti [31]4 years ago

5 0

9514 1404 393

Answer:

24m^14

Step-by-step explanation:

It looks like you want to simplify ...

$1.5m^6(-2m^2)^4=(1.5)(-2)^4(m^6)(m^{2\cdot4})\\\\=(1.5)(16)m^{6+8}=\boxed{24m^{14}}$

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)^c = a^(bc)

(ab)^c = (a^c)(b^c)

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Tessa had $35. She bought a book for $7 and a magazine for $5. Which expression correctly shows the total money Tessa has left?

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4 years ago

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Neil bought pound of cherries. He shares the cherries with friends. Which model

hjlf

Answer:

5 with 1 left

Step-by-step explanation:

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3 years ago

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3 years ago

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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 $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

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3 years ago

Find the scale factor of the figures below from left to right given the volume of each .

Fudgin [204]

<h3>Answer is choice D) 5:3</h3>

Explanation:

The volumes are in the ratio 3625:783 which reduces fully to 125:27 after dividing both parts by the GCF 29.

Both 125 and 27 are perfect cubes because 125 = 5^3 and 27 = 3^3

Therefore, the volume ratio 125:27 turns to the linear scale factor ratio 5:3

If you were to cube both sides of the linear ratio 5:3, you would get the volume ratio 125:27. Cube rooting both sides will have you go in reverse.

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3 years ago