Answer:
the answer is definitely d
Missing info:
Turkey Sizes:
<span>small - 2kg - 3.9kg </span>
<span>medium - 4kg - 5.4kg </span>
<span>large - 5.5kg - 7.2kg </span>
<span>extra large - 7.3kg - 9.2kg </span>
<span>Turkey Prices: </span>
<span>Small / Medium - £5.99 per kg </span>
<span>Large / Extra Large - £4.99 per kg
</span>
3/4 lb x 14 persons = (3*14)/4 = 42/4 = 10.5 lbs
1 lb = 450 grams
10.5 lbs * 450grams/lb = 4,725 grams
1 kg = 1000 grams
<span>4,725 grams * 1kg/1000 grams = 4725 kg / 1000 = 4.725 kg.
4.725 kg is under the Medium size of the Turkey.
Medium sized turkey costs </span>£<span>5.99 per kg.
4.725 kg * </span>£5.99 per kg = £<span>28.30
A turkey would cost </span> £<span>28.30 for 14 people.</span>
Answer:
r = 6
Step-by-step explanation:
V(cyl) = π · r² · h
144π = π · r² · 4
144 = r² · 4
<u>144</u> = r²
4
r =
÷ 
r = 12/2
r = 6
Answer:
The initial mass of the sample was 16 mg.
The mass after 5 weeks will be about 0.0372 mg.
Step-by-step explanation:
We can write an exponential function to model the situation.
Let the initial amount be A. The standard exponential function is given by:

Where r is the rate of growth/decay.
Since the half-life of Palladium-100 is four days, r = 1/2. We will also substitute t/4 for t to to represent one cycle every four days. Therefore:

After 12 days, a sample of Palladium-100 has been reduced to a mass of two milligrams.
Therefore, when x = 12, P(x) = 2. By substitution:

Solve for A. Simplify:

Simplify:

Thus, the initial mass of the sample was:

5 weeks is equivalent to 35 days. Therefore, we can find P(35):

About 0.0372 mg will be left of the original 16 mg sample after 5 weeks.
3a(4a²-5a+12)
12a³-15a²+36a
~Hope this helped!~