(x +7+2i)(x+7-2i) expand and simplify to
x^2 + 14x + 53 <======= take it from here
YOU are in every group of 3 so you need to find the number needed to give two different memebers for 365 days
x C 2 >= 365
trial and error shows x = 28
Area = 60 units2
60 = (b1 + b2)/2 * h
= (b + b+7)/2 * 6
b = 6.5 units
(6.5 + 13.5)/2 = median = 10 units
Please don't just post answers without the method used to reach the solution......it is worthless to do so .
The length of the median of the trapezoid ABCD is 10.
44 * 6 = 264
264 / 8 = ......... erasers in each caddy.
(36 * 8) + (45 * 6) = ....... slices of pizza
28 * 6 = 168
168 / 4 = ....... students in each bus.
\(CBD = 360 - 108 - 47 - 141 = 67 \)
\(\color{brown}\boxed{{\angle}CBD = 67}\)
Plug 5 into the equation:
\(f(5) = 3f(5-2) - 2f(5-1)\)
\(f(5)=3f(3)-2f(4)\)
\(f(3)=3f(1) - 2f(2)\)
\(f(3)=-3 - 6\)
\(f(3)=-9\)
\(f(4)=3f(2) - 2f(3)\)
\(f(4) = 9 +18\)
\(f(4)=27\)
\(f(5)= -27- 54\)
\(\color{brown}\boxed {f(5)=-81}\)
a5 = a1 + 4 d = 20
a100 = a1 + 99d = 305 Now use the sum formula for an arith sequence to sum the 96 terms a5 - a100
96/2 ( 20 + 305) = 15600
Start by dividing both sides by 2.9
1.9x / 1.1x = 1.9655 Now take the x root of both sides
1.9 / 1.1 = 1.96551/x Now LOG both sides
.237361 = 1/x (.29347)
x = 1.2364
Solve for x: 2.9 x 1.72727^x = 5.7
2.9 x 1.72727^x = 29/10 (19/11)^x and 5.7 = 57/10: 29/10 (19/11)^x = 57/10
Multiply both sides by 10/29: (19/11)^x = 57/29
Take the logarithm base 19/11 of both sides: x = log(57/29)/log(19/11)==~1.23642
Hint: 64/27 = 2 10/27
There are 90 4-digit palindromes. 20 of them are divisible by 4 as follows:
(2112, 2332, 2552, 2772, 2992, 4004, 4224, 4444, 4664, 4884, 6116, 6336, 6556, 6776, 6996, 8008, 8228, 8448, 8668, 8888) >>Total = 20 such palindromes.
There are 6 of them as follows:
205 , 214 , 223 , 232 , 241 , 250 , Total = 6
1000 mod360 = 288
288 degrees is in the 4th quadrant
288-360 = -72
it is in the 4th quad so where tan is negative ( not that that matters)
tan(1000) = tan(-72)
n=-72
[AEFGD] = 90 square units