Answer:
2.2 grams
Step-by-step explanation:
We can use the compound decay formula shown below to solve this:

Where
F is the final amount (what we want to find)
P is the present amount (490 grams)
r is the rate of decay, in decimal (28.6% = 28.6/100 = 0.286)
t is the time in minutes (16)
Substituting, we get:

So, after 16 minutes, 2.2355 grams will be remaining.
<em>Rounded to nearest tenth of a gram (1 decimal place) = </em><em>2.2 grams remaining</em>
Answer:
STEM ______ Leaf
2 ______ 0 0 2 4
3 _______ 6 8
4 _______ 2 7
5 ______ 5 7
6 ______ 1 2 3 5
Step-by-step explanation:
Given that data:
20, 24, 65, 36, 47, 55, 62, 20, 22, 63, 38, 42, 57, 61
STEM ______ Leaf
2 ______ 0 0 2 4
3 _______ 6 8
4 _______ 2 7
5 ______ 5 7
6 ______ 1 2 3 5
The unique numbers which starts each value is the stem while the second digit of each unique stem is the leaf.
The answer is 9.
To calculate this, we will use Galileo's square cube law, used to describe the change of the area or the volume of the shape, when their dimensions increase or decrease. The formula used for change of the area is

A₂ - the area after the change,
A₁ - the area before the change,
l₂ - the dimension after the change,
l₁ - the dimension before the change.
It is given:
l₁ = 1 cm
l₂ = 3 cm
Let's use this formula for the mentioned triangle:

Therefore, the area of the triangle was multiplied by 9.
Answer:
Tom: 132 Bob: 396 Keisha: 139
Step-by-step explanation:
132 x 3
132 + 7
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)