Population are always modeled by exponential growth equation given by:
f(x)=a(1+b)^t
where:
a=original number
b=rate of growth
t=time
thus
f(x)=412(1.05)^t
the population after 9 months will be:
f(9)=412(1.05)^9=639.15=639
Answer:
The answer is C: 75
Step-by-step explanation:
The constant is found by multiplying your height and width, so thats really all you do.
25x3=75
15x3=75
3x15=75
25x3=75
75 would be the constant.
y -3 = 2(x-4)
Use distributive property on the right side:
y - 3 = 2x -8
Add 3 to both sides:
y = 2x -5
The answer is A.
Answer: the associative property of addition
Step-by-step explanation: when adding three or more terms, you can combine the middle terms with either the first term or the last term and as you add the results, the sum will be the same either way.
Answer:
Answer for 2nd is option c, for 3rd is option d, for 4th is option e
Step-by-step explanation:
As we know 1 ft.=12 in.
- In ΔABC
∴ The congruent sides are AB and AC respectively
- CB =12 ft. 4 in.=148 in.
- AB=CB =111 in. =9 ft. 3 in.
- AC=CB =111 in. =9 ft. 3 in.
∵ <em>Perimeter of ΔABC</em> =AB+AC+CB
=9 ft. 3 in. + 9 ft. 3 in. +12 ft. 4 in.
=30 ft. 10 in.
2. In ΔDEF
∴ The congruent sides are DE and DF respectively
- DE = 6 ft. 3 in. =75 in.
- DF = 6 ft. 3 in. =75 in.
- Let the length of FE is equal to x
- 0.75FE =DE =DF
- 0.75x = 6 ft. 3 in. =75 in.
- x =100 in. =8 ft. 4 in.
∵ <em>Perimeter of ΔDEF</em> =DE+DF+FE
= 6 ft. 3 in. +6 ft. 3 in. +8 ft. 4 in.
= 20 ft. 10 in.
3. In ΔJKL
∴ The congruent sides are JL and KL respectively
- JK = x+3
- KL =4x-17
- JL =6x-45
- JL≅KL
- 4x-17 =6x-45 . . . . . . . . . . . . . . . . . . . . . . . (1)
- Subracting 4x from both sides from eq 1
- -17 =2x-45
- Adding 45 on both the sides
- 28 =2x
- Dividing by 2 on both sides
- 14 =x
- JK = 14+3 =17
- KL = 4×14-17 =39
- JL = 6×14-45 =39
∵ <em>The dimensions of the ΔJKL are 39,39 and 17.</em>