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USPshnik [31]
3 years ago
12

Let a and b 2 positive real number such that a

Mathematics
1 answer:
lilavasa [31]3 years ago
7 0

Answer:

Proved

Step-by-step explanation:

Given

Integers: a and b

Where a < b

Required

Show that a^n < b^n

We start by writing out the given expression

a < b

Then, take nth root of both sides

a^n < b^n

For instance:

Let a = 2,  b = 2 and n = 4

First

Since 2 < 3

Then

2^4 < 3^4

16 < 81

Proved

<em>This is so for all positive values of a,b and n where a < b</em>

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A rate of 90 kilograms per hour is approximately how many miles per hour? Round to the nearest whole number.
STatiana [176]
I'll assume by kilograms you mean kilometers.

1 kilometre is approximately 0.621 miles, so multiply 90x0.621 to find your conversion.
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3 years ago
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A runner maintains constant acceleration after starting from rest as she runs a distance of 60.0 m. The runner's speed at the en
Svet_ta [14]

Answer:\frac{40}{3}s

Step-by-step explanation:

This is a typical problem of u.a.r.m. ( uniformly accelerated rectilinear motion). Let's take an eye on the u.a.r.m. equations:

v(t) = v_{0} + a.t\\x(t) = x_{0} + v_{0}.t + \frac{1}{2}.a.t^{2}

We know the runner started from rest so we can discard v_{0} as v_{0} = 0

Also we can consider she starts at x_{0} = 0.

All we have to do now is replace these values at our equations to obtain:

v(t) = a.t\\x(t) = \frac{1}{2}.a.t^{2}

As we are told that the acceleration remains constant, we can then find the time it took for her to run the whole distance as follows:

First we note that the ratio between her velocity and the time it takes for her to run a certain distance is constant. This is:

\frac{v(t)}{t} = a ; \forall t

In particular, if we take t=t_{f} where t_{f} is the total time it takes for her to run the whole distance x(t=t_{f})=60m

We know that at this point the velocity is:

v(t=t_{f}) = 9\frac{m}{s}

So that:

a = \frac{v(t=t_{f})}{t_{f}} = \frac{9\frac{m}{s}}{t_{f}}

Now we can replace this in the equation of motion x(t)

x(t=t_{f}) = \frac{1}{2}.a.t^{2} = \frac{1}{2}.\frac{9\frac{m}{s}}{t_{f}}.t_{f}^{2} = 60m

Where we finally find that

t_{f} = \frac{40}{3}s

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3 years ago
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lisabon 2012 [21]

Answer:

its D. 456 in

Step-by-step explanation:

if u multiply 10 x 10 then u get 100. that same square is there twice so if u add it u get 200. then the only one greater than 100 is 456.

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Subtract. Write your answer as a fraction in simplest form. 1/7 - (-4/7)= ?
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Answer:

\frac{1}{7}  - ( -  \frac{4}{7} ) \\  \frac{1}{7}  +  \frac{4}{7}  =  \frac{5}{7}

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