Answer:
Step-by-step explanation:
42+(1/3)
42*(3/3)+(1/3)
(126/3) + 1/3
126/3 + 1/3 = 127/3 miles
4/3 hours
(127/3)/4/3)
(127/3)*(3/4)
= 127/4 or 31.75 miles/hour
It would have been nicer if we were told Graham's distance after 1 hour (when he had travelled 31.75 miles).
3: 18
4: 17
5: 561
6: 17
7: 23
:)
Let's solve your equation step-by-step.
4x−2=0
Step 1: Add 2 to both sides.
4x−2+2=0+2
4x=2
Step 2: Divide both sides by 4.
4x
4
=
2
4
x=
1
2
Answer:
x=
1
2
hope it helps
From the order of operations, we do parens, then multiply/divide, then add/subtract, rewriting each step to keep everything clear:
(14) - 2 - (50 ⋅ 2) <span>⋅ 1 - 1
14 - 2 - 100 </span><span>⋅ 1 - 1 [parens are finished]
14 - 2 - 100 - 1 [multiplication is finished]
-- we do the subtraction left to right, but here are the steps:
12 - 100 - 1
-88 - 1
-89</span>
Answer:
Part a) The height that the liquid forms equals 2.56 feet.
Part b) The number of trucks required are 361.
Step-by-step explanation:
When all the liquid mixture is spilled from the storage tank it will accumulate over the floor space.
Thus we infer
![V_{spilled}=V_{accumulated}](https://tex.z-dn.net/?f=V_%7Bspilled%7D%3DV_%7Baccumulated%7D)
where
is spilled volume of mixture = ![2900m^{3}](https://tex.z-dn.net/?f=2900m%5E%7B3%7D)
is accumulated volume over the floor
Since the floor is smooth thus we conclude that the accumulated volume will form a prism of liquid with height 'h'
Thus
![V_{accumulated}=Area\times h=40000\times h](https://tex.z-dn.net/?f=V_%7Baccumulated%7D%3DArea%5Ctimes%20h%3D40000%5Ctimes%20h)
Equating both the volumes we get
![40000\times h=2900m^{3}\times \frac{35.314ft^{3}}{m^{3}}=102412.533ft^{3}\\\\\therefore h=\frac{102412.533}{40000}=2.56 feet](https://tex.z-dn.net/?f=40000%5Ctimes%20h%3D2900m%5E%7B3%7D%5Ctimes%20%5Cfrac%7B35.314ft%5E%7B3%7D%7D%7Bm%5E%7B3%7D%7D%3D102412.533ft%5E%7B3%7D%5C%5C%5C%5C%5Ctherefore%20h%3D%5Cfrac%7B102412.533%7D%7B40000%7D%3D2.56%20feet)
Thus the height formed by the liquid equals 2.56 feet.
Part b)
We know that the basic relation between density, mass and volume is
![Mass=density\times volume](https://tex.z-dn.net/?f=Mass%3Ddensity%5Ctimes%20volume)
Since it is given that the density of the liquid is 112% the density of water thus we have
![density_{liquid}=1.12\times density_{water}\\\\density_{liquid}=1.12\times 1000=1120kg/m^3](https://tex.z-dn.net/?f=density_%7Bliquid%7D%3D1.12%5Ctimes%20density_%7Bwater%7D%5C%5C%5C%5Cdensity_%7Bliquid%7D%3D1.12%5Ctimes%201000%3D1120kg%2Fm%5E3)
Thus the weight of the volume of liquid is
![mass=2900\times 1120=3248000kg](https://tex.z-dn.net/?f=mass%3D2900%5Ctimes%201120%3D3248000kg)
Now since it is given that 1 truck can carry 9000 kg of liquid thus by proportion the number of trucks need to carry 3248000 kilograms of liquid equals