Just simply divide then multiply by 100
1/6= 0.166667×100=<span>16.6667%
So </span>16.6667% is the answer
if you want it rounded 16.7%
If there are 4 marbles left over each time, then we can forget about them for now.
So the question is, what is the smallest number than can be divided into 6,7 and 8?
the numbers have only one non-1 divisor in common: both 6 and 8 are divisible by 3.
so for our purposes we can "delete" one 2 and ask:
what is the smallest number than can be divided into 3,7 and 8 ?
There are no more divisors in common, so we just have to multiply them: 3*7*8=21*8=168
and the 4 marbles "extra"? We add them to this sum.
the the smallest possible number in the box is 168+4=172.
Answer:
<u>Given:</u>
- DC ║ AB
- CM = MB as M is midpoint of BC
i) <u>Since DN and BC are transversals, we have:</u>
- ∠DCM ≅ ∠NBM and
- ∠CDM ≅ ∠BNM as alternate interior angles
<u>As two angles and one side is congruent, the triangles are also congruent:</u>
- ΔDCM ≅ ΔNBM (according to AAC postulate)
So their areas are same.
ii)
<u>The quadrilateral has area of:</u>
- A(ADCB) = A(ADMB) + A(DCM)
<u>And the triangle has area of:</u>
- A(ADN) = A(ADMB) + A(NBM)
Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.
Answer:
The pebble travelled 149.25 feet vertically
Step-by-step explanation:
Answer:
By exterior angle theorem, we have;
∠DBE = ∠H + ∠HEB = ∠ECD = ∠H + ∠HDC
∴ ∠H + ∠HEB = ∠H + ∠HDC
By addition property of equality, we have
∠HEB = ∠HDC
∠H = ∠H by reflexive property
∴ ΔHCD ~ ΔHEB by Angle Angle AA similarity postulate
∴ HE/HD = EB/DC, by the definition of similarity
Therefore, by cross multiplication, we have;
HE × DC = EB × HD
Therefore, by commutative property of multiplication, we have;
HE × DC = HD × EB
Step-by-step explanation:
