The problem statement gives a relation between the amount removed from one bag and the amount removed from the other. It asks for the amount remaining in each bag. Thus, there are several choices for variables in this problem, some choices resulting in more complicated equations than others.
Let's do it this way: let x represent the amount remaining in bag 1. Then the amount removed from bag 1 is (100-x). The amount remaining in bag 2 is 2x, so the amount removed from that bag is (100-2x). The problem statement tells us the relationship between amounts removed:
... (100 -x) = 3(100 -2x)
... 100 -x -3(100 -2x) = 0 . . . . . . subtract the right side
... 5x -200 = 0 . . . . . . . . . . . . . . eliminate parentheses and collect terms
... x -40 = 0 . . . . . . . . . . . . . . . . .divide by 5
... x = 40 . . . . . . . . . . . . . . . . . . . add 40
- 40 kg is left in the first bag
- 80 kg is left in the second bag
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<u>Check</u>
The amount removed from the first bag is 60 kg. The amount removed from the second is 20 kg. The amount removed from the first bag is 3 times the amount removed from the second bag, as described.
Answer:
A triangle with side lengths 4 cm, 5 cm, and 15 cm
Step-by-step explanation:
In order to form a triangle with given side lengths, sum of any two sides must always be greater than the third side.
Answer:
D.$315
Step-by-step explanation:
$450=100%
100%-20%-10%=70%
$450×70%=$450×70/100
=$315
Answer:
404 cm³ Anyway... Look down here for my explanation.
Step-by-step explanation:
Let's Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth on your paper. A right-angled triangle is formed too. The Length of side to the water-surface is 5 cm, the hospot is 7 cm.
We Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7)
44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8°
The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram.
Shaded area 88.8/360*area of circle - ½*7*788.8°
= 88.8/360*π*7² - 24.5*sin 88.8°
13.5 cm²
(using area of ∆ = ½.a.b.sin C for the triangle)
Volume of water = cross-sectional area * length
13.5 * 30 cm³
404 cm³
Answer:
all the sides and corners have lengths may i have brainly
Step-by-step explanation:
