A way to work this type of problem is to set up two 'railroad track' computations - one assuming that the iron III is limiting and the other assuming that the chlorine is limiting. Whichever results in the smaller amount of Iron III chloride being produced is the correct calculation...

The derivative of f(x)) can be worked out to be
f' = 1/(x+e)2 - 2 x /(x +e)3
Setting this equal to zero and multiplying by (x +e)2 results in
1 - 2x /(x+e) = 0. This equation is easily solved to get x =...

The Heron formula can be used.
A = sqrt( s (s-a)( s -b) (s-c)) where s is the semi perimeter. s = (4 + 6 + 8)/ 2= 9
A = sqrt( 9 x 5 x 3 x 1) = sqrt(135)

The key to problems like this is that rates add.
Let p = the filling rate for the pipe and h = the filling rate for the hose.
Then h + p = 1/3 (that is - a rate of one third of a pool per hour)
...

Define P(A) = probability that car A will start (the prior for car A starting)
P(B) = probability that car B will start (the prior for car B starting)
The probability that B starts given that A does not...

The answer is almost. The function A(d) is monotone increasing. However, it has a range restriction. A negative value for A (your age) does not make a lot of sense. (At least for dates more than 9 months before you were born.) Consequently, the inverse function...

The locus is a single point called the in center. The in center is the point where all three angle bisectors intersect.
The in center is the center of the inscribed circle. The inscribed circle is tangent to all three sides.

The formulas that you want for parts c) and d) are
u = (1/2) ε0 Emax2 and
u = (1/2) (1/μ0 Bmax2
where ε0 = 8.854 E-12 and μ0 = 12.578 E-7
Using u = 5.6E-6
Emax = 1124.7 V/m and
...

The fact that one root of P(x) is i, implies that -i is also a root. (complex roots always occur as complex conjugate pairs). Since i and -i are roots, P(x) has a factor of (x-i)(x+i) = x2 +1.
Synthetic division or long division can be used...

This looks to be a standard textbook type problem. The total impedance is given by
Z = sqrt( R2 +(XL-XC)2) = 444.6 Ω
The max emf is given by Emax = Imax Z = 400.1 V.
The maximum voltage across the capacity is Vc = Imax XC = ...

Assuming it is log7(7 √7) that is wanted:
the answer is (3/2) log7(7) = 3/2

The reaction shown is a reduction of an aldehyde to an alcohol. This is typically done with a reducing agent such as
NaBH4. The aldehyde is reduced, but the atom within the aldehyde that is gaining the electrons (being reduced) is usually thought of as the being the carbon...

The expression on the left hand side can be factored so:
(2 cos(θ) -1) ( 2 cos(θ) +3) = 0
Thus either 2 cos(θ) -1 =0 or 2 cos(θ) +3 =0
but there are no solutions for 2 cos(θ)...

The standard method for determining the empirical formula given the concentration by mass is the following:
1) consider a sample of 100 g. Then work out the mass of each element. For this problem, the mass of C would be 24.24 g, for H, 4.04 g, for Cl, 71.72...

Call the center of the circle point F. Then m∠HFJ = 180 - 101 = 79. The angle HFJ is a central angle. Therefore the minor arc HJ also has measure 79. The major arc HJ has measure 360 - 79 = 281.

The apothem, a, of a regular pentagon is the distance from the center of the pentagon to the midpoint of an edge.
Trigonometry can be used to show that a = (s/2) /tan(pi/10), where s is the length of a side. This is an exact result.
The area, A, of...

The simplest way to set this up is to note that 1 - .028 = .972 So
A(t) = A(0) .972t Where t is number of hours.
Since the formula listed is in e format this equation can be rewritten as
A(t) = A(0) exp [ ln(...

There are two possible answers depending on how the 9 ft is interpreted. If the 9 ft is the distance along the ground between Charles and the mirror, the calculation is:
6/9 = h/63 and h = 42.
If the 9 ft is the...

This is a fairly interesting problem with several elements.
The first is to note that 8 cos(θ) - 6 sin(θ) is equal to 10 cos(θ + arctan(3/4) ) .
This is a standard transformation that can be checked by using the cosine of a sum formula.
The arctan(3/4) is just a phase shift...

The maximum possible volume would be for the case that the rectangular prism is a cube. In that case each of the six sides would have an area of 724 / 6 = 120.66 cm2 . The edge would be sqrt(120.66) = 10.98 cm. The volume would be the cube of this: V = 1325...