4 for part A
17 for part B
Ok so divide 39 by 4 which is 9.75 then mulitiply that by 3 to get your answer of $29.25
Answer:
<u>Solution</u><u> </u><u>given</u><u>:</u>
<u>one</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u> </u><u>must</u><u> </u><u>be</u><u> </u><u>less</u><u> </u><u>than</u><u> </u><u>the</u><u> </u><u>sum</u><u> </u><u>of</u><u> </u><u>two</u><u> </u><u>other</u><u> </u><u>side</u><u> </u><u>.</u>
<u>by</u><u> </u><u>making</u><u> </u><u>these</u><u> </u><u>sense</u><u>:</u>
3<10+20
10<20+3
20<10+3not true
these three side are not possible.
again
10<20+25
20<25+10
25<20+10
these three side are true so
required side are:
Side 1:10
Side 2:25
Side 3:20
Answer:
Since length is three times, hence it is 6 inches.
let the width be x.
then, length= 3x
perimeter= 16 inches.
perimeter of a rectangle = 2( L+B)
2(3x+x)=16.
3x+x= 16÷2
4x= 8
x= 2
since length=3x
hence, 3×2= 6 inches
Let the number of girls be x and the number of boys be y
The number of boys is the same as the number of girls.
x = y
The number of boys is twice the number of girls when 8 girls left.
2(x - 8) = y
x = y ---------------- (1)
2(x - 8) = y -------- (2)
Sub (1) into (2)
2(y - 8) = y
2y - 16 = y
2y - y = 16
y = 16
There were 16 boys and 16 girls initially.
16 + 16 = 32
There were a total of 32 students in the class initially.
