Answer:
2) Add 21 to both sides
Step-by-step explanation:
When solving 
for 
, our goal to isolate such that we have set equal to something.
Therefore, we want to start by adding 21 to both sides. This leaves us with 
and we are one step closer to isolating .
Answer:
1
Step-by-step explanation:
Answer:
<h2><u><em>
a²+2ab+b²-c²</em></u></h2>
Step-by-step explanation:
Solve:
(a+b+c) (a+b-c)=
(a²+ab-ac+ab+b²-bc+ac+bc-c²)=
a²+ab-ac+ab+b²-bc+ac+bc-c²=
a²+2ab+0ac+b²+0bc-c²=
a²+2ab+b²-c²
Answer:
In 1981, the Australian humpback whale population was 350
Po = Initial population = 350
rate of increase = 14% annually
P(t) = Po*(1.14)^t
P(t) = 350*(1.14)^t
Where
t = number of years that have passed since 1981
Year 2000
2000 - 1981 = 19 years
P(19) = 350*(1.14)^19
P(19) = 350*12.055
P(19) = 4219.49
P(19) ≈ 4219
Year 2018
2018 - 1981 = 37 years
P(37) = 350*(1.14)^37
P(37) = 350*127.4909
P(37) = 44621.84
P(37) ≈ 44622
There would be about 44622 humpback whales in the year 2018
9514 1404 393
Answer:
√35 +3√7 -6 . . . square units
Step-by-step explanation:
The area can be figured a number of ways. The figure can be divided into parts, and the areas of those parts added.
Or, the area of the enclosing rectangle can be found, and the rectangle at upper right that is not shaded can be subtracted from that. We choose the latter.
The overall width is the sum of the given partial widths:
width = (√7 -2) + (2) = √7
Then the area of the bounding rectangle is ...
A = LW = (√5 +3)(√7) = √35 +3√7
The area of the upper right empty-space is ...
A = LW = (3)(2) = 6
Then The area of the shaded figure is ...
√35 +3√7 -6
