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4 years ago

I need help ASAP PLEASE!

Mathematics

1 answer:

Mandarinka [93]4 years ago

7 0

Answer:

Only numbers we can think of would be 0 and 1,where cube of the numbers would be equal to the number itself.

Step-by-step explanation:

But in the case of positive numbers the constraints of the question can never be met. So, we'll have to use negative numbers as

(-ve number) ^ (even number) is always a postive number and (-ve number)^(odd number) is always a negative number.

Eg:- (-2)^3= -2 *-2 *-2= -8

(-2)^2= 4

Since, negative sign can never be associated with 0. Eliminate 0 as a possible answer.

Only option left is -1. Lets test if -1 is correct or not.

(-1)^3= -1*-1*-1= -1 i.e the number itself.

(-1)^2=-1*-1=1 which isn't equal to -1.

Therefore, -1 is the solution to this problem.

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Select the correct answer. Consider this absolute value function. f(x)= |x-5| How can function f be written as a piecewise funct

faltersainse [42]

Option (c) is correct according to me I think it’s helpful

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3 years ago

Read 2 more answers

Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w

Alinara [238K]

Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100× $(\frac{1}{2} )^{3.322}$ .

Step-by-step explanation:

The equation for half life is of the form

A = A₀× $(\frac{1}{2} )^{\frac{t}{h} }$ .........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ $\frac{t}{h}$ and and the equation (2) can be written as

Percentage remaining after 2 half lives is

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{2 }$

However if the half life of Technetium-99m is 6 hours then we have for one day

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{2 *2}$

Therefore an expression that represents the number of days until only 10% remains is

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{\frac{d}{h} }$ = 10 %

$(\frac{1}{2} )^{\frac{d}{h} }$ = $\frac{1}{10}$

= ㏑ $(\frac{1}{2} )^{\frac{d}{h} }$ = ㏑( $\frac{1}{10}$ )

= $\frac{d}{h}$ ×㏑ $(\frac{1}{2} )$ = ㏑( $\frac{1}{10}$ )

$\frac{d}{h}$ = $\frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )}$ = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100× $(\frac{1}{2} )^{3.322}$

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4 years ago

It takes 45 minutes to drive to the nearest bowling alley taking city streets going 30 miles per hour

zmey [24]

Answer:

The time it will take to get to the bowling alley while driving at 65 mph on the freeway is approximately 20.8 minutes

Step-by-step explanation:

The given parameters are;

The time it takes through the streets to drive to the nearest bowling alley = 45 minutes

The speed of driving through the streets = 30 miles per hour (mph)

The speed of driving through the freeway to the bowling alley = 65 mph

From the speed of driving through the streets and the time taken we have;

The formula for speed, S, is S = Distance/Time

Therefore, 30 mph = Distance/45 minutes

Where, 1 hour = 60 minutes, 45 minutes = 1/60×45 hour = 3/4 hour

30 mph = Distance/45 minutes = Distance/(3/4 hour)

30 mph = Distance/(3/4 hour)

Distance = 30 mph × 3/4 hour = 22.5 miles

Given that the distance through the freeway and through the streets are the same, we have from speed = Distance/Time;

Time = Distance/Speed

The speed through the freeway = 65 mph while the distance remain 22.5 miles

Therefore;

Time = 22.5 miles/(65 mph) ≈ 0.346 hour ≈ 20.8 minutes

The time it will take to get to the bowling alley while driving at 65 mph on the freeway is approximately 20.8 minutes.

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4 years ago

What is the product of 2 irrational numbers.

soldier1979 [14.2K]

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3 years ago

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Alina [70]

Multiply price of the jersey by number of jerseys needed.

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3 years ago