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3 years ago

Rewrite each of the expressions below in a simpler form using exponents.

Mathematics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

6y^4

Step-by-step explanation:

Tell me if you need an explanation.

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The ice cream shop offers 31 flavors. you order a double scoop cone how many different ways can the clerk put the ice cream on t

Lynna [10]

Answer:

465

Step-by-step explanation:

Since we are looking for the the number of combinations to have 2 different flavors in the ice cream, we can use the formula for combination.

$_{n}C_{k}=\dfrac{n!}{k!(n-k)!}$

Our variables are:

n = 31

k = 2

$_{31}C_{2}=\dfrac{31!}{2!(31-2)!}$

$_{31}C_{2}=\dfrac{31!}{2!29!}$

$_{31}C_{2}=465$

The number of possible combinations of having 2 different flavors out of 31 different flavors is 465.

6 0

3 years ago

Which product will be positive?<br> -6 × 4<br> -6 × (-7)<br> -3 × 8<br> 6 × (-4)

masya89 [10]

Answer:

-6x(-7)( negative and negative = positive)

Step-by-step explanation:

-6x4= -24

-6x(-7) = 42

-3x8= -24

6x(-4) = -24

7 0

3 years ago

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For what value(s) of k will the function y=6x^2-8x+k have

liubo4ka [24]

Think the answer is a

8 0

3 years ago

The volume of a cone-shaped funnel is 603 cubic centimeters. The radius of the funnel is 8 centimeters. Find the height of the f

yarga [219]

Answer: 9

Step-by-step explanation:

Volume of cone formula is: pi r^2 h/3

So pi x 8^2 h/3 = 603

H/3 = 2.9991

h = 9 (rounded)

4 0

3 years ago

A sample of a radioactive substance decayed to 96.5% of its original amount after a year. (Round your answers to two decimal pla

jek_recluse [69]

Answer:

The half-life of the substance is 19.47 years.

Step-by-step explanation:

The equation for the amount of substance remaining is given by the following equation:

$Q(t) = Q(0)e^{-rt]$

In which Q(t) is the amount remaining after t years, Q(0) is the initial amount and r is the rate that this amount decreases.

A sample of a radioactive substance decayed to 96.5% of its original amount after a year.

This means that $Q(1) = 0.965Q(0)$

We use this to find r. So

$Q(t) = Q(0)e^{-rt]$

$0.965Q(0) = Q(0)e^{-r]$

$e^{-r} = 0.965$

$\ln{e^{-r}} = \ln{0.965}$

$-r = =0.0356$

$r = 0.0356$

$Q(t) = Q(0)e^{-0.0356t}$

(a) What is the half-life of the substance?

This is t when Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0356t}$

$0.5Q(0) = Q(0)e^{-0.0356t]}$

$e^{-0.0356t} = 0.5$

$\ln{e^{-0.0356t}} = \ln{0.5}$

$-0.0356t = \ln{0.5}$

$t = -\frac{\ln{0.05}}{0.0356}$

$t = 19.47$

The half-life of the substance is 19.47 years.

8 0

4 years ago

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