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beks73 [17]
3 years ago
10

Please help me solve this priblem

Mathematics
2 answers:
Over [174]3 years ago
5 0
It would be c. 12 inches

H = V/lw

The Length * Width = 450
and the volume is 5400
So 5400/450 = 12

Hope that helps!
polet [3.4K]3 years ago
5 0
It is 12 inches because 5400 divided by 450= 12
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A large rectangular office measures 9 yards by 15 yards. What is the area of the office?
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What is the decimal of 64/100
DENIUS [597]

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0.64

Step-by-step explanation:

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For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

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5 0
2 years ago
For each of the -3.3 trials, and in the same sequence in which you entered the freezing points, enter your calculated values for
dedylja [7]

Answer:

The morality of the solution is calculated as 0.859 m. We are required to determine the freezing point depression constant of pure water. The freezing point depression of the solution is given as

T_{f}  = T_{p}  -T_{s} = K_{f} * (morality of solution)

T_{p} and T_{s} are the freezing points of the pure solvent (water, 0°C) andK_{f} = freezing point depression constant of water. Therefore,

T_{f}  = K_{f} *(0.859 m)

=====> (0°C) – (3.00°C) = K_{f}  *(0.859 m)

=====> -3.00°C = K_{f}  *(0.859 m)

Ignore the negative sign (since K_{f} is positive) and get

K_{f} = (3.00°C) / (0.859 m) = 3.492°C/m

The freezing point depression constant of the solvent is 3.5°C kg/mole  

3.5 K.kg/mole (temperature differences are the same in Celsius and Kelvin scales).

3 0
3 years ago
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