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3 years ago

How do I graph a solution set for the inequality 300+5x>550?

1 answer:

3 0

First to solve the inequality
300 + 5x > 550
5x > 550 - 300
5x > 250
x > 50

Therefor the answers are all numbers bigger than 50

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John is w years old. His sister is 16 years old and his brother is twice johns age all three siblings have a total age of 31

Kisachek [45]

Answer:

Step-by-step explanation:

John's age = w

His brother's age = 2*w = 2w

His sister's age = 16 years

Total age of all siblings = 31

w + 16 + 2w = 31

w + 2w + 16 = 31

Combine like terms

3w + 16 = 31

Subtract 16 from both sides

3w = 31 - 16

3w = 15

w = 15/3

w = 5

$\boxed{\text{John's age = 5 years }}$

His brother's age = 2w

= 2*5

= 10

$\boxed{\text{John's brother's age = 10 years}}$

6 0

3 years ago

Least common multiple <br> 14,6

Vlada [557]

Answer:

Find the prime factorization of 14

14 = 2 × 7

Find the prime factorization of 6

6 = 2 × 3

Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:

LCM = 2 × 3 × 7

LCM = 42

6 0

3 years ago

Read 2 more answers

Helpppppppppp me pleaseeeee

Vlad1618 [11]

Answer:

A = 16 13unit^2 + 9unit^2 = 5 √10 = 15.8 = 16

Step-by-step explanation:

For Pythagoras we square route

= x^2 = 13^13 + (one of the closer squares for x elimination)

x= 169 + 81 169 + 100 169 + 49 169+64 169 + 121

x = 250 or 269 or 218 or 233 or 290

x= 5√10 or 15.8 or √269 269 or √218 2.8 or √233 or 2.33 ( or √290 = ( √48+2) or 2.9

6 0

3 years ago

24.8 divided by 175.

WITCHER [35]

24.8 divided by 175 is 0.1417142857142857 or you can round it to 0.142

5 0

3 years ago

Read 2 more answers

Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

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

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

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

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

7 0

3 years ago