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Nady [450]
1 year ago
9

michael bought a new ipad for $450. Alexander bought the same ipad at a 12% discount. how much did alexander pay

Mathematics
2 answers:
Ahat [919]1 year ago
6 0

Answer: $396

Step-by-step explanation:

Michael bought the ipad for $450

Alexander bought if at 12% discount.

12% discount of the price = 12% of 450 = $54

Therefore , Alexander will buy it for

$450 - $54 = $396

lara [203]1 year ago
3 0
396$ because it’s right
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Answer:

x=8

Step-by-step explanation:

The figure is shown in the attached figure.

We have, LK║OM. We need to find the value of x.

It is based on side-splitter theorem. If a line is parallel to third side of a triangle, then the points that intersects the two lines, it divides proportionally.

Using it in given problem. So,

\dfrac{\text{NK}}{\text{KM}}=\dfrac{\text{NL}}{\text{LO}}

Nk = x+2, KM = x-3, NL = x and LO = x-4

So,

\dfrac{x+2}{x-3}=\dfrac{x}{x-4}

Cross multiplying, we get :

(x+2)(x-4)=x(x-3)\\\\x^2-4x+2x-8=x^2-3x\\\\-2x+3x=8\\\\x=8

So, the value of x is 8.

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Equation: t = 25 + 8.25w Aaliyah needs $135.75 for a television. How many weeks will it take her to save enough money to buy a t
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An investor is interested in purchasing an apartment building containing six apartments. The current owner provides the followin
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b) 2.7632

Step-by-step explanation:

To find the mean, we multiply each value by it's probability. So

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What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re
sashaice [31]

Answer:

4.0921 is the logarithm of the equilibrium constant.

Step-by-step explanation:

Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

Ag^+(aq) + e^-\rightarrow Ag(s); E° = 0.80 V

Iron having negative value of reduction potential .So ,that means that it will loose electron easily and get oxidized.Hence, will be at anode.

E^{o}_{cell}=Reduction potential of cathode - Reduction potential of anode

E^{o}_{cell}=E^{o}_c-E^{o}_a

=0.80 V-(-0.41 V)=1.21 V

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Net reaction: Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)

n = 2

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

\Delta G^o=-nfE^o_{cell}

and,

\Delta G^o=-RT\ln K_{eq}

Equating these two equations, we get:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 1.21 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}

\ln K_{eq}=9.3478

\log K_{eq}=\frac{9.3478}{2.303}=4.0921

4.0921 is the logarithm of the equilibrium constant.

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1 year ago
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