We have the form
(x - h)^2 (y - k)^2
______ + ________ = 1
a^2 b^2
a =(1/2) length of major axis = (1/2)(20) = 10........a^2 = 100
b = (1/2) length of mi,or axis = (1/2) ( 16) = 8 .....b^2 =64
(5,3) =( h, k)
So we have
(x -5)^2 (y - 3)^2
_______ + ________ = 1
100 64
Let's say Mary has 60 marbles and Jolie has 100 marbles (since it would work in this context). If we add them together, there are 160 marbles in total.
Because we're looking for Jolie's percentage, we need to find the value of x for:
100/160 = x/100
which we know is 62.5.
Therefore 62.5% of the marbles in the bag are Joli's.
Hope this helped!
Caffeine :)
The area of ABC = (1/2)(AC)(BC) sin ACB = (AC))BC) sin ABC / 2
And the area of triangle FCE = (1/2) (AC/3) ( 2BC/3) sin ACB = (AC) (BC) sin ACB / 9
So [ FCE ] / [ ABC ] = 2/9 ⇒ [FCE] = (2/9) [ ABC]
And each of the white areas = 2/9 area of ABC
So their total area = 3(2/9)area of ABC = 2/3 area of ABC
So area of DEF / area of ABC = [ area of ABC - 2/3 area of ABC ] / area of ABC = 1/3
100% (Total DNA) - 48% (A-T Pair) = 52% (C-G Pair)
T=70 +0.4(52) = 90.8 C
wheres the link.
Thanks!
PQRS is a rectangle with PQ=16 cm and QR=12 cm. If ABCD is a rhombus formed PQRS, what is the radius of the circle inscribed in rhombus ABCD?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
AO = 6 cm BO = 8 cm
∠ABO = tan-1(6 / 8)
Radius EO = sin(∠ABO) * BO = 4.8 cm
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ or
AB = sqrt(AO2 + BO2) = 10 cm
Radius EO = (AO * BO) / AB = 4.8 cm
Cross-multiply
59 [ ( n + 1)! - 3(n - 1)! ] = 53 [ (n + 1)! +3 (n - 1)! ]
(59 - 53)(n + 1)! = (59*3 + 53*3) ( n -1)!
6(n + 1)! = 336 ( n -1)!
(n + 1)! / ( n -1)! = 336 / 6
(n + 1)! / ( n -1)! = 56 [ note : (n + 1)! / (n - 1)! = (n + 1) n ]
(n + 1) n = 56
8 * 7 = 56
n = 7
2 ^(16^x) = 16^(2^x) take the log of each side
log 2 ^(16^x) = log 16^(2^x) and we can write
(16^x) log 2 = (2^x) log 16
(16^x) / (2^x) = log 16/log 2
(16/2)^x = log 16/log 2
8^x = log 2^4 / log 2
8^x = (4log 2) / log 2
8^x = 4
(2^3)^x = 2^2
2^(3x) = 2^2 equate exponents
3x = 2
x = 2/3
See heureka's excellent answer , here :
https://web2.0calc.com/questions/stuck-on-a-geometry-question
f(x) must be a quadratic just like g(x)
So
(ax^2 + bx + c)^2 - 11(ax^2 + bx + c) + 30 =
a^2 x^4 + 2 a b x^3 + 2 a c x^2 + b^2 x^2 + 2 b c x + c^2 - 11ax^2 - 11bx - 11c + 30
a must =1
And
2ab must = 2(1)b = -14 ⇒ b = -7
(2ac + b^2 - 11a) = 62
2c + 49 - 11 = 62
2c + 38 = 62
2c = 24
c =12
Check
2bc -11b = -91 ??? c^2 - 11c + 30 = 42 ???
2(-7)(12) -11(-7) 12^2 - 11(12) + 30 =
-168 + 77 144 - 132 + 30 =
-91 42
f(x) = x^2 - 7x + 12
f(7) = 7^2 - 7(7) + 12 = 12
Diagonal length = sqrt(2 * 400) = 28.3
If the baby smiled in 30% of the pictures (aka 12 photos), the answer is simply 12/30%=40
Proof: 40*0.3=12 so it works
A square is partly within a circle and partly outside it, with two sides tangent to the circle as shown in the figure.
If the radius of the circle is 4 units. Find the area of the square correct to two decimal places.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
r = 4
CO = sqrt(2r2)
AC = CO + r
AB = sqrt(AC2/2)
[ABCD] = AB2 ≈ 46.63 u2
Thank u very much!
THX, IWTH!!! .....Here's a little easier way
First term = 88
Common difference = -3
We have
88 - 3(n -1) = -17 where -17 is the nth term
88 - 3n + 3 = -17
-3n + 91 = -17
-3n = -108
n = -108/-3 = 36th term
So.....35 terms appear before -17.....just as IWTH found !!!!
You have
30%:12
10%:4
10*10%:4*10
100%:40
So, the photographer took 40 pics
(88, 85, 82, 79, 76, 73, 70, 67, 64, 61, 58, 55, 52, 49, 46, 43, 40, 37, 34, 31, 28, 25, 22, 19, 16, 13, 10, 7, 4, 1, -2, -5, -8, -11, -14, -17)
So 35 terms appear BEFORE -17.
Let C = (0,0) let B = (8,6)
Then BC = sqrt ( 8^2 + 6^2) = sqrt (100) = 10
Draw OM perpendicular to BC
This will form two right triangles OBM and OCM
OB = 8 OM = r BM = sqrt ( OB^2 -r^2) = sqrt ( 8^2 - r^2)
OC = 6 OM = r MC = sqrt (OC^2 - OM^2) = sqrt ( 6^2 - r^2)
BM + MC = BC
sqrt ( 8^2 - r^2) + sqrt ( 6^2 -r^2) = 10
sqrt (64 - r^2) + sqrt (36 - r^2) =10
sqrt (64 -r^2) = 10 - sqrt (36 -r^2) square both sides
64 -r^2 = 100 - 20sqrt (36 - r^2) + 36 -r^2
64 = 136 - 20 sqrt (36 - r^2)
-72 = -20 sqrt (36 - r^2)
72/20 = sqrt (36 -r^2)
18/5 =sqrt (36 - r^2) square both sides
324/25 = 36 - r^2
-576/25 = -r^2
r^2 =576/25 take the positive root
r = 24/5 cm
You can draw a picture if you need too
Both men walk 40 yards in opposite directions and then 30 yards to left.
The two men are standing on the ends of two triangles, if you use the pythagoras theorem,
we know that the distance between both of them is 100 yards.
30%=12 Pictures
30/3=10% 12/3=4 Pictures
10%=4 Pictures
10*10=100% 4*10=40 Pictures So, the photographer took 40 pictures.
Y=3 X=-3 3+(-3)=0 9+(-18)=-9
To simplify this, let them start at (0,0)
Let them first walk east-west
Then one is at (40,0) and the other is at (-40,0)
Then...when each turn left and walk for 30 yars each.....one is at ( 40,30) and the other is at ( -40, -30)
Using the distance formula
D =sqrt [ ( 40 - - 40)^2 + ( 30 - -30)^2 ]=
sqrt [ 80^2 + 60^2 ] =
sqrt [ 6400 + 3600 ] =
sqrt [ 10000 ] =
100 yds apart
a) Inverse of f
Get x by itself, "swap" x and y
y =2x + 1
y-1 =2x
(y -1) / 2 = x
(x - 1) / 2 = y = the inverse
b) f(g) we are putting g into f
f(g) = 2( 3x - 2) + 1 = 6x - 4 + 1 = 6x - 3
c) g(f).....just the other way around.....we put f into g
g(f) = 3( 2x + 1) - 2 = 6x + 3 - 2 = 6x + 1
d) 2f(g) =3g (f)
2 (6x - 3) = 3(6x + 1)
12x - 6 = 18x + 3
12x - 18x = 3 + 6
-6x = 9
x = 9/-6 = -3/2
what do you need help with @CPhill?
