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Vitek1552 [10]
2 years ago
8

Is x greater than, less than, or equal to 43° ?

Mathematics
1 answer:
solong [7]2 years ago
5 0

Answer:

X is equal

Step-by-step explanation:

look at the angle

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Find a number You can add to 6274 so you have to regroup ones,tens, and Hundreds
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6274+1000= 7274 ?
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A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig
Vaselesa [24]

Answer:

a) Depth changing rate of change is 0.24m/min, When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of 0.17m/min, When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them \frac{3.5}{10} =\frac{a}{h}, then a=0.35h.

a) As we have that volume is V=\frac{1}{2} 2ahL=ahL, then V=(0.35h^{2})L, so we can derivate it \frac{dV}{dt}=2(0.35h)L\frac{dh}{dt} due to the chain rule, then we clean this expression for \frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt} and compute with the knowns \frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min, is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is 2a=0.7h, we can derivate it and obtain \frac{da}{dt}=0.7\frac{dh}{dt}  =0.7*0.24m/min=0.17m/min The width of the top of the water is changing, When the water is 6 meters deep at this rate

8 0
3 years ago
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8 0
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A 3mL unit dose of levalbuterol HCl contains 0.63mg of active ingredient. If 2 unit doses are used, how many mg of active ingred
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Answer:  0.42 mg

Step-by-step explanation:

Given : A 3mL unit dose of levalbuterol HCl contains 0.63mg of active ingredient.

Then , the amount of active ingredient in 1 ml dose of levalbuterol HCl = \dfrac{0.63}{3}

=\dfrac{63}{300}=\dfrac{21}{100}=0.21\ mg

Hence, the amount of active ingredient in 2 ml dose of levalbuterol HCl will be :-

2\times0.21=0.42\ mg

Hence, the amount of active ingredient in 2 ml dose of levalbuterol HCl =0.42 mg

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