Answer:
Widow's Share = Rs 500.05
Son's share = Rs 1400.14
Step-by-step explanation:
Property = 4000.40
Widow get share = 0.125
So, Share of Widow = 0.125 * 4000.40
Widow's Share = Rs 500.05
Remaining Property = 4000.40 - 500.05
Remaining Property = 3500.35
Son's share = 0.4 * 3500.35
Son's share = Rs 1400.14
Answer: 60
Step-by-step explanation:
Let the side lengths of the rectangular pan be m and n. It follows that (m-2) (n-2)=
So, since haf of the brownie pieces are in the interior. This gives 2 (m-2) (n-2) =mn
mn- 2m - 2n- 4 = 0
Then Adding 8 to both sides and applying, we obtain (m-2) (n-2) =8
Since now, m and n are both positive, we obtain (m,n) = (5,12), (6,8) (up to ordering). By inspection, 5. 12 = 60
which maximizes the number of brownies in total which is the greatest number.
Hope that helped! =)
The content of care package would be a board game and several snack packs.
Given is the weight of board game = 4 pounds.
The weight of one snack pack =
The total weight would be less than 25 pounds.
Solving for part A :
Let 'x' represents the number of snack packs.
So the weight of all snack packs would be 0.5x pounds.
Now weight of care pack = weight of board game + weight of all snack packs.
weight of care pack = 4 + 0.5x
But total weight must be less than 25 pounds.
So the inequality would be: 4 + 0.5x < 25
Solving for part B :
Solving the inequality 4 + 0.5x < 25
⇒ 4 + 0.5x - 4 < 25 - 4
⇒ 0.5x < 21
⇒ x < 42
Solving for part C :
Since x < 42 and x is an integer number, so x = 41.
She can include maximum of 41 snack packs in the care package.
Answer:
4/5
Step-by-step explanation:
cos B = 3/5
cos Ф = adjacent/hypothesis
hypothesis ² = adjacent² + opposite²
5²= 3²+ o²
25 = 9+o²
25-9=o²
16 = o²
√16 = o = 4
Sin B = opposite/hypothesis = 4/5