5y=4x+10 /:5
y=4/5 x+2
y=mx+k
m₁*m₂=-1
4/5 *m=-1
m=-5/4
y=-5/4 x+k
-5/4*(-15)+k=8
75/4 + k=8
k=32/4 - 75/4
k=-53/4=-13.25
y=-5/4 x - 53/4
Do 15.5×12
then you will know how much ribbon you need. Hope that helps!
Answer:
- The smallest area the field could be is 6,400 m²
- The largest area the field could be is 8,250 m²
Step-by-step explanation:
Given;
smallest possible length of the international soccer field, L₀ = 100 m
smallest possible breadth of the international soccer field, B₀ = 64 m
Largest possible length of the international soccer field, L₁ = 110 m
Largest possible breadth of the international soccer field, B₁ = 75 m
Area of a rectangle is given by;
A = L x B
The smallest area the field could be is calculated as;
A₀ = L₀ x B₀
A₀ = 100 m x 64 m
A₀ = 6,400 m²
The largest area the field could be is calculated as;
A₁ = L₁ x B₁
A₁ = 110 m x 75 m
A₁ = 8,250 m²
The answer is

÷ 2 =

÷
If

÷

=

, then:

÷

=
Now, 2 can be canceled out from

:
Answer:
2 
Step-by-step explanation:
for question a)
It would be easier if you converted these to improper fractions (currently it is in the form of a proper fraction)
to convert to improper fraction, you would multiply the denominator (bottom no. of fraction) by the whole number , then add that with the numerator (top no. of fraction) and put the whole thing over the denominator.
E.g.
7
= 7x4 + 1 = 29
Thus, improper fraction would be 
do same thing for other one
thus , 
now,
- 
however, you can't do this as the denominators are not the same, so multiply them by a number which would give the a common denominator (20 would be the best option for common denominator)
= 
convert back to mixed number
2 