Make 2 eqn, let a be apples and o be oranges:
1. 5a + 4o = 10
2. 5a + 5o = 11
set up an elimination:
1. minus 2.
5a + 4o = 10
- 5a + 5o = 11
____________
o = 1
sub “o” into one of the eqn
5a + 5(1) = 11
5a = 11-5
5a = 6
a = 6/5
a = 1.2
therefore, an orange costs $1.00 and an apple costs $1.20
I am so so so sorry I do not know the answer but I might later in the shool year
Including Jamiz, there are 14 girls in the class. If this if half of the total number of pupils (as half are girls, the other half being boys) the total number of students is 14 × 2 = 28 students
20%.
20,000 times 0.8 is 16,000, so the decreased value that was lost was 20% of the original price.
Answer:
1.8m^2 approx
Step-by-step explanation:
Given data
P1= 1400N
A1=0.5m^2
P2=5000 N
A2=??
Let us apply the formula to calculate the Area A2
P1/A1= P2/A2
substitute
1400/0.5= 5000/A2
cross multiply
1400*A2= 5000*0.5
1400*A2= 2500
A2= 2500/1400
A2= 1.78
Hence the Area is 1.8m^2 approx
