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3 years ago

Find the range of each function given the domain; g(m)=m^2;(-2,0,2)

Mathematics

1 answer:

Anettt [7]3 years ago

4 0

$g(-2)=(-2)^2=4\\
g(0)=0^2=0\\
g(2)=2^2=4$

The range is $\{0,4\}$

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What is the value of 6(2b-4) when b=5.

trapecia [35]

<h3><u>3</u><u>6</u> is the answer.</h3>

Step-by-step explanation:

Put value of b as given and using BODMAS rule,

$6(2b - 4)$

$= > 6((2 \times 5)-4)$

$= > 6(10 - 4)$

$= > 6 \times 6$

$= > 36$

Henve, value of of the given expression is $36$

7 0

1 year ago

Read 2 more answers

Helps, also explain how to do it I want to learn how and stop using brainy

Nutka1998 [239]

Answer:

259

Step-by-step explanation:

USE PEMDAS

Parenthesis

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5 0

3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago

What is the greatest common factor of the following monomials: 12g^5h^4 g^5h^2

muminat

Answer:

g^5h^2

Step-by-step explanation:

12g^5h^4, g^5h^2

This is one way of doing it. Break down every number and every variable into a product of the simplest factors. Then see how many of each factor appear in both monomials.

12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h

g^5h^2 = g * g * g * g * g * h * h

So far you see every single prime factor of each monomial.

Now I will mark the ones that are present in both. Those are the common factors.

12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h

g^5h^2 = g * g * g * g * g * h * h

The greatest common factor is the product of all the factors that appear in both monomials.

GCF = g * g * g * g * g * h * h = g^5h^2

6 0

2 years ago

a survey was done Of 600 shoppers at a grocery store to determine if they like a new flavor Of potato chip. Of the 600 shoppers,

olasank [31]

Answer:

63%

225 buyers

Step-by-step explanation:

To find the percent take the number that liked it over the total

376/600

Change to decimal form

.62666666

Change to percent by multiplying by 100

62.66666666%

The nearest percent is 63%

37.5 % would but it then multiply by the number of shoppers

37.5 % ( 600)

Change to decimal form

.375 * 600

225

7 0

3 years ago