Answer:
19. 11
21. 119
Step-by-step explanation:
19.
(-5)² - [4(-3 ∙ 2 + 4)² + 3] + 5 =
= (-5)² - [4(-6 + 4)² + 3] + 5
= (-5)² - [4(-2)² + 3] + 5
= (-5)² - [4(4) + 3] + 5
= (-5)² - [16 + 3] + 5
= 25 - 19 + 5
= 6 + 5
= 11
21.
5 - 8[6 - (3 ∙ 2 - 8 + 2|4 ÷ -2 + (-3)| - 4) - 7 · 2] - 3² · (-2) =
= 5 - 8[6 - (3 ∙ 2 - 8 + 2|-2 + (-3)| - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - (3 ∙ 2 - 8 + 2|-5| - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - (3 ∙ 2 - 8 + 2(5) - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - (6 - 8 + 10 - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - (-2 + 10 - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - (8 - 4) - 7 · 2] - 3² · (-2)
= 5 - 8[6 - 4 - 7 · 2] - 3² · (-2)
= 5 - 8[6 - 4 - 14] - 3² · (-2)
= 5 - 8[2 - 14] - 3² · (-2)
= 5 - 8[-12] - 3² · (-2)
= 5 - (-96) - 9 · (-2)
= 5 + 96 + 18
= 101 + 18
= 119
Answer:
225
Step-by-step explanation:
Because of 45 time 5
Y=7x+9 here’s your answer
Answer:If you would like to know what will the approximate population be after 3 years, you can calculate this using the following steps:
an initial population ... 298 quail
an annual rate ... 8%
an exponential function to model the quail population:
f = 298(1+8%)^t = 298(1+8/100)^t
f ... quail population
t ... time (years)
t = 3 years
f = 298(1+8/100)^t = 298(1.08)^3 = 375.4 quail
375.4 quail after 3 years.
