Kramer collected = 225000
Elaine collected = 5000
Newman collected = 45000
Solution:
Given Jerry collected 45000.
Kramer collected = 5 times more than Jerry collected
= 45000 × 5
= 225000
Kramer collected = 225000
Elaine collected = 9 times less than Jerry collected
= 45000 ÷ 9
= 5000
Elaine collected = 5000
Newman collected = same as Jerry collected
= 45000
Newman collected = 45000
Hence, Kramer collected = 225000
Elaine collected = 5000
Newman collected = 45000
Answer:
B; The area of the triangle is 12 square centimeters
Step-by-step explanation:
In this question, we are asked to calculate the area of a pendant in the shape of a regular hexagon which is divided into a number of congruent triangles with each triangle having an area of 2 square centimeters.
The key to answering this question is simply knowing the number of triangles that will be present in the regular hexagon.
Mathematically, since the triangles are congruent and have the same area, this means the triangles must be equal, meaning we have 6 equilateral triangles.
To get the area of the regular hexagon here, we simply multiply the area of one of the triangles by 6.
Mathematically, this is 6 * 2 = 12 square centimeters
Answer:
show us the graphs to be able to tell you which one
Step-by-step explanation:
Answer:
AB ≈ 15.7 cm, BC ≈ 18.7 cm
Step-by-step explanation:
(1)
Using the Cosine rule in Δ ABD
AB² = 12.4² + 16.5² - (2 × 12.4 × 16.5 × cos64° )
= 153.76 + 272.25 - (409.2 cos64° )
= 426.01 - 179.38
= 246.63 ( take the square root of both sides )
AB =
≈ 15.7 cm ( to 1 dec. place )
(2)
Calculate ∠ BCD in Δ BCD
∠ BCD = 180° - (53 + 95)° ← angle sum in triangle
∠ BCD = 180° - 148° = 32°
Using the Sine rule in Δ BCD
=
=
( cross- multiply )
BC × sin32° = 12.4 × sin53° ( divide both sides by sin32° )
BC =
≈ 18.7 cm ( to 1 dec. place )
Answer:
Step-by-step explanation:
In ΔXYZ and ΔZWX,
∠WXZ ≅ ∠YZX [Given]
∠XZW ≅ ∠ZXY [Given]
XZ ≅ XZ [Reflexive property]
Since, two corresponding angles and the included side of two triangles are equal,
ΔXYZ ≅ ΔZWX [ASA postulate of congruence]