Answer:
you didn't add a question
Step-by-step explanation:

- Mass of the truck (m) = 30000 Kg
- Velocity (v) = 25 m/s
- Let the momentum be p.
- We know, momentum = mass × velocity
- Therefore, the momentum of the truck (p)
- = mv
- = (30000 × 25) Kg m/s
- = 750000 Kg m/s
<u>Answers</u><u>:</u>
<u>Variable </u><u>:</u><u> </u><u>m,</u><u> </u><u>v,</u><u> </u><u>p</u>
<u>Equation </u><u>:</u><u> </u><u>p </u><u>=</u><u> </u><u>mv</u>
<u>Answer:</u><u> </u><u>The momentum of the truck is 750000 Kg m/s.</u>
Hope you could understand.
If you have any query, feel free to ask.
Answer:
a. 277.3 m²
b. $307.86
Step-by-step explanation:
a. Area of the wall to be painted = area of rectangle - area of triangle
= L × W + ½×a×b×sin θ
L = 17 m
W = 14 m
a = 8.7 m
b = 9.5 m
θ = 72°
Plug in the values into the equation
Area of the wall = (17×14) + (½*8.7×9.5×sin 72)
Area of the wall = 238 + 39.3024105
Area of the wall to paint ≈ 277.3 m²
b. 20 liters of paint of 1 container cost $21.99
If 1 liter of paint covers 1m², therefore,
277.3 m² will need = 277.3 × 1 = 277.3 liters of paint.
20 liters = 1 container of paint
277.3 liters = 277.3/20 = 13.865 ≈ 14 containers
1 container = $21.99
14 containers = 21.99 × 14 = $307.86
If 3x>96, dividing both sides by 3 we get that x>32, therefore only E is an acceptable answer
Answer:
Step-by-step explanation:
4) ΔSTW ≅ ΔBFN . So, corresponding parts of congruent triangles are congruent.
a) BN = SW d) m∠W = m∠N
BN = 9 cm m∠W = 82°
b) TW = FN e) m∠B = m∠S
TW = 14 cm m∠B = 67°
c) BF = ST f) m∠B + m∠N + m∠F = 180°
BF = 17 cm 67 + 82 + m∠F = 180
149 + m∠F = 180
m∠F = 180 - 149
m∠F = 31°
5) ΔUVW ≅ ΔTSR
UV = TS
12x - 7 = 53
12x = 53+7
12x = 60
x = 60/12
x = 5
UW =TR
3z +14 = 50
3z = 50 - 14
3z = 36
z = 36/3
z = 12
SR =VW
5y - 33 = 57
5y = 57 + 33
5y = 90
y = 90/5
y = 18
7) ΔPHS ≅ ΔCNF
∠C = ∠P
4z - 32 = 36
4z = 36 + 32
4z = 68
z = 68/4
z = 17
∠H = ∠N
6x - 29 = 115
6x = 115 + 29
6x = 144
x = 144/6
x = 24
∠P + ∠H + ∠S = 180 {Angle sum property of triangle}
36 +115 + ∠S = 180
151 + ∠S = 180
∠S = 180 - 151
∠S = 29°
∠F = ∠S
3y - 1 = 29
3y = 29 + 1
3y = 30
y = 30/3
y = 10
8) ΔDEF ≅ ΔJKL
DE = 18 ; EF = 23
DF = 9x - 23
JL= 7x- 11
DF = JL {Corresponding parts of congruent triangles}
9x - 23 = 7x - 11
9x - 7x - 23 = -11
2x - 23 = -11
2x = -11 + 23
2x = 12
x = 12/2
x = 6
JK = DE {Corresponding parts of congruent triangles}
3y - 21 = 18
3y = 18 + 21
3y = 39
y = 39/3
y = 13