Let x = shorter side
Let 2x-4 = longer side
The ratio between the two sides of rectangle B must be the same as that of rectangle A since they are similar. We set up an equation based on this ratio 12/16 (which equals 3/4).
x/(2x-4) = 3/4
I understand this to be the arc cosine of the cosine of 16pi/5. Since cosine and arc cosine are inverse functions, we get back the original value as our answer. That would be 16pi/5.
1. AB ≅ AE (given)
2. BC ≅ ED (given)
3. AC ≅ AD (addition postulate - this states that the sums of equal quantities are equal)
If f(12) = 4, then we will use x=4 for the first graph. When we substitute x for 4, we get y = 3f(12) - 5 = 3*4-5 = 12-5 = 7. Therefore, the point (4,7) exists on the graph of y = 3f(x+8) - 5
For the second graph, we would substitute x for 1/9. Then we would get:
Your first derivative didn't use a correct form of differentiation.
f(t) = t(4-t2) = 4t - t3
f'(t) = 4 - 3t2
We set the derivative equal to zero
4 - 3t2 = 0
4 = 3t2
4/3 = t2
t = +/- 2√3/3
This is a parabolic function. Therefore, the minimum or maximum will occur at the point where t = -b/2a.
b = 63
a = -16
t = -63/-32 = 63/32
The height at t=63/32 would be:
h(63/32) = -16(63/32)2 + 63(63/32) + 4 = -3969/64 +...
To differentiate this function, we must use the product rule of differentiation:
y = f(x) * g(x) * h(x)
y' = f'(x)*g(x)*h(x) + f(x)*g'(x)*h(x) + f(x)*g(x)*h'(x) =
ex(csc x) + xex(csc x) - xex(csc x)(cot x)
C = present amount of chlorine
C0 = initial amount of chlorine
k = decay constant
t = time
C = C0ekt
at t=0, C = 2.5. At t=1, C = 2.2
2.2 = 2.5ek
ek = .88
k = ln .88 = -.1278
C(t) = 2...
A. The cost of producing widget number 31 should be the cost of producing 31 widgets minus the cost of producing 30 widgets which is:
C(31) - C(30) = (5000 + 40*31 - 0.02*312) - (5000 + 40*30 - 0.02*302) = 6220.78 - 6182 =
B. Marginal Cost is defined...
The molar mass of an element or compound is equal to the atomic mass of the element or compound except that is measured in grams. For example, an atom of helium has a mass of 4 amu. Therefore one mole of helium would have a mass of 4 grams.
Phosphorous has an atomic mass of 31 amu...
A 200 kilogram chocolate with 13% fat content would have 200 * .13 = 26 kilograms of fat
let x = mass of chocolate with 10% fat content
let y = mass of chocolate with 40% fat content
x + y = 200
.1x + .4y = 26
Multiply the second equation...
Let x = quantity of 10% fat-content chocolate in kilograms
Let y = quantity of 60% fat-content chocolate in kilograms
The two must add up to 100 kilograms of chocolate.
The fat content of the two must add up to 55 kilograms
x + y = 100
This is the same truth value as p V s because the negation of a negation always means the original statement much like -(-8) is the same as 8.
let x = calories burned by running per hour
let x - 230 = calories burned by riding a bike per hour
Joe's two hour run plus warm up = 2x + 150
Tommy's 4 hour bike ride = 4(x - 230)
2x + 150 = 4x - 920
9x4 - 28x2 + 3 = 0
Since the c value is positive, the sign of both factors must be the same. Since there is a minus sign in front of the b value, both factors must have a minus sign.
(9x2 - 1)(x2 - 3) = 0
9x2 - 1 = 0
Since water slows down the boat going upstream & speeds up the boat going downstream by the same speed of 3 mph, the average speed of the boat relative to the water would be the same as the average speed of the boat without the water current. That would (72 miles + 72 miles)/18 hours which...
Sodium Phosphate: Na3PO4
Zinc Nitrate: Zn(No3)2
Barium Bromate: Ba(BrO3)2
Iron (II) perchlorate: Fe(ClO4)2
cobalt (II) hydrogen carbonate: Co(HCO3)2 -----------this should be called cobalt bicarbonate
Hydroiodic acid: HI
Chloric Acid: HClO3
Nitrous Acid: HNO2
H2CO3: Carbonic acid
HClO4: Perchloric acid
CH3COOH: acetic acid
F = Five dollar bills
T = Ten dollar bills
F = 3T
5F + 10T = 675
5(3T) + 10T = 675
15T + 10T = 675
25T = 675
T = 27
F = 3* 27 = 81
There are 27 ten dollar bills
M stands for moles/liter.
1) The atomic mass of sucrose is 342 daltons. This means the mass of one mole of sucrose is 342 grams. To make a 2 molar solution of sucrose, you must have 2 moles of sucrose in 1 liter of solution. 2 moles of sucrose would have a mass of 684 grams...