Answer:
Greatest number of flowers that can be used in a bouquet is 8.
Step-by-step explanation:
In this question greatest number of flowers used in a bouquet will be decided by the "Greatest Common Factor" of the numbers of the flowers given.
So, factors of 16 = 1×2×2×2×2
Factors of 8 = 1×2×2×2
Factors of 32 = 1×2×2×2×2×2
Common factors = 1×2×2×2
Greatest Common Factor of these numbers will be = 1×2×2×2 = 8.
Therefore, greatest number of flowers that could be in a bouquet is 8.
Answer:
Step-by-step explanation
let's use g for golf and b for batting cages
So let's start with Sylvester
he plays 5 rounds of mini golf so 5g
and he takes 4 turns in the batting cages so 4b
and he pays 60 dollars for this
SO his equation is 5g+4b=60
Now onto Lin
3g for 3 rounds of golf
6b for 6 turns in the batting cages
He pays 45 dollars
SO his equation is 3g+6b=45
Both equations:
5g+4b=60
3g+6b=45
Now you need to cancel out one variable so you can multiply the first equation by 3 and the second one by -5
15g+12b=120
-15g-30b=-225
Now the g will cancel out when you add both equations
-18b=-105
b=105/18 which is about 5.83 dollars
Now plug in 105/18 into any of the original equations and solve for g
The three inside angles of a triangle need to equal 180 degrees.
In a right triangle one of the angles is 90 degrees.
This means the other two angles need to equal 180-90 90 degrees.
You are given one of the angles of 52 degrees, so subtract that from 90 to find the third angle.
90 - 52 = 38 degrees.
The answer is A.
For a, the starting value (intercept) is 6599. The depreciation rate 12%/year. This is where the function y = 6599(1-(rate)^t) comes into play. 6599 is the starting price, rate is the 12%, and t is the time in years. So the equation is: y = 6599(0.88)^t. For part b, simply plug 2.5 in for t which results in $4,794. For part c, same deal, plug in 20 for t, which results in $512. For part d, you have to take the anti-log where 2000 = 6599(0.88)^t... divide both sides by 6599... 2000/6599 = (0.88)^t. Which is approximately 8.5 years.
