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1 year ago

What is the first step to solve for n in the following equation? 3n - 7 = 30

Mathematics

2 answers:

sergejj [24]1 year ago

5 0

Answer:

Simplifying

3n + 7 = 30

Reorder the terms:

7 + 3n = 30

Solving

7 + 3n = 30

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '-7' to each side of the equation.

7 + -7 + 3n = 30 + -7

Combine like terms: 7 + -7 = 0

0 + 3n = 30 + -7

3n = 30 + -7

Combine like terms: 30 + -7 = 23

3n = 23

Divide each side by '3'.

n = 7.666666667

Simplifying

n = 7.666666667

Step-by-step explanation:

astra-53 [7]1 year ago

3 0

Answer:

3n - 7 = 30

3n = 30 + 7

3n = 37

n = 37/3

n = 12 1/3

Step-by-step explanation:

hmu if you need more help! :))

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lidiya [134]

0.35 cups/hour

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12 cups/34 hours= 0.35 cups/hour
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2 years ago

Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original salary?

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Her original salary was $500

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1 year ago

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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

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

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

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

$2Ag^+(aq) + 2e^-\rightarrow 2Ag(s)$ ; E° = 0.80 V

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

10 x 3 tens - unit form and standard form

vfiekz [6]

<span>10X3 tens in unit form is written:
10*3 tens = 30 tens = 300 units

10X3 tens in standard form:
10 x 3 tens = 10 x 30 = 300</span>

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2 years ago

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A circle fits inside a semicircle of diameter 24mm

patriot [66]

Answer:

The shaded are is $113\ mm^{2}$

Step-by-step explanation:

we know that

The shaded area is equal to the area of the semicircle minus the area of the circle inside the semicircle

step 1

Find the area of semicircle

The area of semicircle is equal to

$A=\frac{1}{2}\pi r^{2}$

where

$r=24/2=12\ mm$ ----> the radius is half the diameter

substitute

$A=\frac{1}{2}\pi(12)^{2}=72\pi\ mm^{2}$

step 2

Find the area of the circle inside the semicircle

The area of circle is equal to

$A=\pi r^{2}$

where

$r=12/2=6\ mm$ ---> the radius of circle inside is half the radius of semicircle

substitute

$A=\pi(6)^{2}=36\pi\ mm^{2}$

step 3

Find the shaded area

$72\pi\ mm^{2}-36\pi\ mm^{2}=36\pi\ mm^{2}$

assume

$\pi=3.14$

$36(3.14)=113.04\ mm^{2}$

3 significant figures is

$113\ mm^{2}$

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1 year ago