Answer:
(5d⁴ − 8)²
Step-by-step explanation:
I hope this is helpful!
Answer:
Q1
<u>The exponential growth model for frogs:</u>
F- number of frogs, x- number of years, 1.22 - growth factor
<u>Calculations</u>:
- F(5) = 100(1.22)⁵ = 270
- a) F(10) = 100(1.22)¹⁰ = 730
- b) F(20) = 100(1.22)²⁰ = 5335
<em>All numbers rounded </em>
Q2
<u>The exponential growth model for bacteria:</u>
B- number of bacteria, x- number of hours, 1.8 - growth factor
<u>Calculations:</u>
- a) B(5) = 10(1.8)⁵ = 189
- b) B(24) = 10(1.8)²⁴ = 13382588
- c) B(168) = 10(1.8)¹⁶⁸ = 7.68 * 10⁴³
<em>All numbers rounded </em>
Q3.
<u>The exponential growth model for fish:</u>
F- number of fish, x- number of months, 1.02 - growth factor
<u>Calculations:</u>
- a) F(12) = 821(1.02)¹² = 1041
- b) F(120) = 821(1.02)¹²⁰ = 8838
<em>All numbers rounded </em>
7x- y = 17 is the standard form
<h3>
Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
<h3>
Step-by-step explanation:</h3>
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
_____
<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy