Answer:
This is a long answer. I got 82.5
Step-by-step explanation:
Area of isoceles triangle is

where b is the base and h is height.
Let draw a altuide going through Point D that split side FE into 2 equal lines. Let call that point that is equidistant from FE, H.
Since it is a altitude,it forms a right angle. So angle H=90.
Angle H is equidistant from F and E so
FH=11
EH=11.
The height is still unknown.
We can use pythagorean theorem to find side h but we need to know the slanted side or side DF to use the theorem.
Using triangle DFH, we know that angle H is 90 and angle F is 34. So using triangle interior rule,Angle D equal 56.
- We know side FH=11
- We know Angle D equal 56
- We are trying to find side DF
- We know angle H equal 90
We can use law of sines to find side DF

Plug in the numbers

sin of 90 =1 so

Side df is about 13.3 inches.
Since we know our slanted side is 13.3 we can set up our pythagorean theorem equation,





is approximately 7.5 so dh=7.5 approximately.
Now using base times height times 1/2 multiply them out


Answer:
# of 20cent stamps = 8
Step-by-step explanation:
Let the number 20cent stamps be = x
Let the number 60 cents stamps be = y
Total number of stamps => 20 = x + y ----------- ( 1 )
Total cost of the stamps = 8.80
That is, 0.20x + 0.60y = 8.80 ------------- ( 2 )
( 2 ) => 20x + 60y = 880
=> x + 3y = 44
=> x = 44 - 3y
Substitute x in ( 1 ) => x + y = 20
44 - 3y + y = 20
- 2y = 20 - 44
- 2y = - 24
2y = 24
y = 12
Substitute y = 12 in ( 1 ) => x + y = 20
x + 12 = 20
x = 8
Answer:
225 necklaces.
Step-by-step explanation:
Assuming that they have the necessary amount of straws to match the beads then we can use the following simple equation to solve this problem.
t = b/12 ... where t is the total amount of necklaces and b is the number of beads you have
t = 2700 / 12
225 necklaces.
For 225 necklaces you would also need 1125 straws since each necklace requires 5 straws to be fully made. Therefore if you have this amount you should not run into any problems.
Answer:
D for the first one A for the second