Given that a room is shaped like a golden rectangle, and the length is 29 ft with the ratio of golden rectangle being (1+√5):2, thus the width of the room will be:
ratio of golden triangle=(length if the room)/(width of the room)
let the width be x
thus plugging the values in the expression we get:
29/x=(1+√5)/2
solving for x we get:
x/29=2/(1+√5)
thus
x=(29×2)/(1+√5)
answer is:
x=58/(1+√5)
or
byrationalizing the denominator by multiplying both the numerator and the denominator by (1-√5)
58/(1+√5)×(1-√5)/(1-√5)
=[58(1-√5)]/1-5
=(58√5-58)/4
Answer:
2
Step-by-step explanation:
First thing you gotta do is find the common denominator.
Do this by listing factors of the denominators:
5- 5, 10, 15, 20, 25, 30
6- 6, 12, 18, 24, 30, 36
5 and 6 both have 30 in common so that's the new denominator for all 3 of the terms.
What you do to the bottom you must do to the top.
Since you multiply 5 by 6 to get 30, you must multiply 2 and 4 by 6 as well.
Since you multiply 6 by 5 to get 30, then multiply 5 by 5 as well.
The new equation is:
+
+
=
Now add the numerators together
The solution is
. Now simplify. 61 is prime, so you cannot change the fraction How many times does 30 go into 61? Two times. This gives us 2
.
You do 5 multiple by 2 + 2. 4 multiple by 3 + 3. 12 multiple by 15 is 180
Answer:
make 10 copies and each will be sold for 200dollars
Answer:
the cost of running the boarding
house for 600 students is N61,000
Step-by-step explanation:
Let C represents cost
K1 represents first constant
K2 represents second constant
C= k1+k2n
3500 = k1 + 25 k2............. Eqn(1)
6000= k1 + 50 k2 .............. Eqn(2)
Subtract eqn(1) from eqn(2)
2500= 25k2
K2= 2500/25
K2= 100
To get k1 from eqn(1)
3500 = k1 + 25 k2
Substitute the value of k2
3500 = k1 + 25 (100)
3500= k1 +2500
K1= 3500- 2500
K1= 1000
The equation connecting them;
C= 1000+ 100n
The cost of running the boarding
house for 600 students is
n= 600
C= 1000+ 100(600)
C= 1000+60000
C= N 61,0000