AE = 5*sqrt(2), CE = 3*sqrt(3)
The graph has period \(\frac{2 \pi}{3}.\) The period of y = a tan bx is \(\frac{\pi}{b},\) so \(b = \frac{3}{2}.\) The graph is then of the form \(y = a \tan \left( \frac{3x}{2} \right).\)
Since the graph passes through \(\left( \frac{\pi}{6}, 2 \right)\), \(2 = a \tan \frac{\pi}{4} = a.\)Therefore, \(ab = 2 \cdot \frac{3}{2} = \boxed{3}.\)
You can use the formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) to solve this. The s variable represents the semiperimeter which is: 2 + 5 + 6 = 13 , half of that is 7.5.
\(A = \sqrt{7.5(7.5-2)(7.5-5)(7.5-6)} = \sqrt{7.5\times20.625} = \sqrt{154.6875} = \boxed{\frac{15}{4}\sqrt{11}}\)
ab = 4.
Actually, 4 works too
Using a calculator, 1, 2, and 3 work, but not 4, so the answer is 4.
Why don't you try it...
Start with the number 1. Can you make it bigger? Can you make that number bigger? Repeat this, and you will find your answer...
a0 = 4
a3 = 6
So
a2 + 2a1 = 6 (1)
a1 + 2a0 = a2 → a1 + 2*4 = a2 → a1 - a2 = -8 (2)
Add (1) and (2)
3a1 = -2
a1 = -2/3
a2 = 6 - 2a1 = 6 - (2)(-2/3) = 22/3
a4 = a3 + 2a2
a4 = 6 + 2(22/3) = 62/3
a5 = 62/3 + 2(6) = 98/3
0 1 2 3 4 5
4 -2/3 22/3 6 62/3 98/3
Ok, thanks you for your help!
7 ingredients
7c1 + 7c2 + 7c3 + 7c4 + 7c5 + 7c6 + 7c7
7 + 21 + 35 + 35 + 21 + 7 + 1
= 127 combos of ingredients
1 type of bread: 5 options
5 * 127 = 635 sandwiches
2 types of bread: 5c2 = 10
10 * 127 = 1270 sandwiches
635 + 1270 = 1905 sandwiches
Let us suppose that N tickets were sold
Then
[ .75N*20 + .25N* 15 ] / N =
N [ 15 + 3.75 ] / N =
15 + 3.75 =
$18.75 = average selling price
k = 7/3.
Problem: Calculate \arccos \sqrt{\cfrac{1+\sqrt{\cfrac{1-\sqrt{\cfrac{1-\sqrt{\cfrac{1+\cfrac{\sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}}
As usual, the output of an inverse trig function should be in radians.
Answer: 4/9*pi
Let's suppose that such integers exist.....we can write
5 ( 5a + 3b) = 1
(5a + 3b) = 1 / 5
But (5a + 3b) must be an integer, not a fraction
So......no such integers exist
BD = 5.
See my answer under your original post
Let the cost of the bill = C
Will paid (1/3)C
Mike and Sue together paid (2/3) C
Mike paid (2/7) (2/3)C
Sue paid (5/7)(2/3)C
And
What Mike paid + $6 = What Sue paid .....so....
(2/7) (2/3)C + 6 = (5/7)(2/3)C
(4/21)C + 6 = (10/21)C
(10/21)C - (4/21)C = 6
(6/21)C = 6
C = 6(21/6) = $ 21
Proof
Will paid (1/3)(21) =$7
Mike paid (2/7) (2/3) (21) = $4
Sue paid $10
=> for 24 cookies he need = 1 4/3 cups of sugar
i.e 7/3 cups
1 cookie he need = 7/3 / 24
= 7/3 * 1/24
for 18 cookies he need = 7/(3 * 24 4) * 18 * 3
= 7/4 cups
i.e 1 3/4 cups
No. of pieces Parker got candy = 11
Each of his friends got these times candy as he did.
So each of his friends got candy = 3 * 11 = 33
No. of pieces of candy get Parker and his friends = 11 + 33 + 33 + 33 = 110
Each of his friends got 3*11 = 33 pieces
11 + 3(33) =
110 altogether
a1 = 2^0
a2 = 2^1 * 1 = 2^1
a3 = 2^2 * 2^1 = 2^3
a4 = 2^3 * 2^3 = 2^6
a5 = 2^4 * 2^6 = 2^10
a6 = 2^5 * 2^10 = 2^15
.......
The nth term is given by 2 ^ ( n * (n-1) / 2)
a23 = 2 ^ (23 * 22 / 2) = 2^(23 * 11) = 2^253 ⇒ p = 253
expected number of defectives = probability of getting a defective * total number of calculators
total calculators = 6 * 12 = 72
P(defect) = 2/12 = 1/6
expected defectives = 1/6 * 72 = 12
Both triangles are isosceles
Angle AED = 80°
Angle ADE = 180 - 2(80) = 20°
Angle DEC = 2ADE = 2 * 20 = 40°
Angle BEC = 180 - 80 - 40 = 60°
Angle EBC = (180 - 60) / 2 = 120 / 2 = 60°
f(2) implies that
sqrt (x + 1) = 2 → x = 3
f(2) = 1 / 3