Answer:
Step-by-step explanation:
$7.45 = 100g
x = 3kg( 3 × 1000) = 3000g
∵ $7.45 = 100g
x = 3000g
(100 × x) = (7.45 × 3000)
100x = 22350
100x/100 = 22350/100
x = $223.5
If you let the width be x, then the length will be 2x + 3. The perimeter P is
twice the width plus twice the length, i.e.,
P = 2*x + 2*(2x + 3) = 2x + 4x + 6 = 6x + 6.
So if the perimeter is 66 cm then 66 = 6x + 6 ---> 6x = 60 ---> x = 10 cm.,
and so if the perimeter is at most 66 cm., the width can be at most 10 cm..
Answer:
None of the above.
Step-by-step explanation:Yes, all of the sides match, which would make it an equilateral polygon. However, it is not an equiangular polygon because not all of the angels are the same. In this case, since the angles do not mach and only the sides do, this is not a regular polygon. This is what leads me to believe that is none of them. All of the sides must be congruent and all interior angles must also be congruent.
Answer/Step-by-step explanation:
1. Side CD and side DG meet at endpoint D to form <4. Therefore, the sides of <4 are:
Side CD and side DG.
2. Vertex of <2 is the endpoint at which two sides meet to form <2.
Vertex of <2 is D.
3. Another name for <3 is <EDG
4. <5 is less than 90°. Therefore, <5 can be classified as an acute angle.
5. <CDE is less than 180° but greater than 90°. Therefore, <CDE is classified as an obtuse angle.
6. m<5 = 42°
m<1 = 117°
m<CDF = ?
m<5 + m<1 = m<CDF (angle addition postulate)
42° + 117° = m<CDF (Substitution)
159° = m<CDF
m<CDF = 159°
7. m<3 = 73°
m<FDE = ?
m<FDG = right angle = 90°
m<3 + m<FDE = m<FDG (Angle addition postulate)
73° + m<FDE = 90° (Substitution)
73° + m<FDE - 73° = 90° - 73°
m<FDE = 17°
