![jbug6780](/avatars/2154.jpg)
jbug6780
11.03.2020 •
Mathematics
What are the solutions to the inequality (x-3)(x+5)less than or equal to 0
Solved
Show answers
More tips
- G Goods and services What Useful Foods Can You Buy at Supermarkets?...
- G Goods and services How to Choose the Best Publishing House for Your Children s Book: Tips and Recommendations...
- S Science and Technology When do we change our clocks?...
- A Animals and plants How to Teach Your Parrot to Talk?...
- F Family and Home How to Remove Fading from Clothes: Tips and Tricks...
- F Food and Cooking How to Make Polendwitsa at Home?...
- F Family and Home Parents or Environment: Who Has the Most Influence on a Child s Upbringing?...
- P Philosophy Unbelievable stories of encounters with otherworldly forces...
- L Leisure and Entertainment How to Choose the Perfect Gift for Men on February 23rd?...
- H Health and Medicine How to Treat Whooping Cough in Children?...
Answers on questions: Mathematics
- M Mathematics 1. Find the surface area of the pyramid....
- M Mathematics 2 Solve for 1: S Help...
- B Biology The body will not get oxygen if what system fails...
- M Mathematics What is the probability of pulling a 8,9,or 10 of hearts in a deck of cards...
- G Geography Which country borders Venezuela to the east? O Brazil O Colombia O Ecuador O Guyana...
- M Mathematics Ski resorts are interested in the average age that children take their first Ski and Snowboard lessons they need this information to optimally plan their ski classes match the vocabulary...
- C Chemistry Selenium-60 Selenium-85 # of protons # of neutrons # of electrons...
Ответ:
-5 _> x _< 3
Step-by-step explanation:
(x-3)(x+5)less than or equal to 0
(x -3)(x+5) <_ 0
x-3 _< 0 or x + 5_< 0
x _< 3 or x _< -5
Therefore
-5 _> x _< 3
I hope this was helpful, please rate as brainliest
Ответ:
k = 1
P(t) = (t + 20)²
Step-by-step explanation:
P' = k√P = k P⁰•⁵
To solve for k,
P' = 10 rabbits/month
P = 100 rabbits
10 = k √100
10 = 10k
k = 1
To solve for P(t)
P' = dP/dt
(dP/dt) = kP⁰•⁵
dP/P⁰•⁵ = k dt
P⁻⁰•⁵ dP = k dt
∫ P⁻⁰•⁵ dP = ∫ k dt
2P⁰•⁵ = kt + c
At t = 0, P = 100
2(100)⁰•⁵ = 0 + c
2 × 10 = c
c = 20
2P⁰•⁵ = kt + c
2P⁰•⁵ = kt + 20
Recall, k = 1
2P⁰•⁵ = t + 20
P⁰•⁵ = (t + 20)
P(t) = (t + 20)²