Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
Answer:
y=21.14
Step-by-step explanation:
x=6 because after 6 weeks
y= 3.27(6)+ 1.52
y=19.62 + 1.52
y=21.14
Answer:
Year 10 Interactive Maths - Second Edition
Problem Solving
Linear equations are often used to solve practical problems that have an unknown quantity. We use a suitable pronumeral to represent the unknown quantity, translate the information given in the problem into an equation, and then solve the equation using the skills acquired earlier in this chapter.
Example 11
If a number is increased by 8, the result is 25. Find the number.
Solution:
Let x be the number. Increasing x by 8 gives x + 8, which we are told is 25. Therefore, x + 8 = 25. Subtract 8 from both sides to find x = 17. So, the number is 17.
Step-by-step explanation:
I hope this helps!!!
Answer:
z^1+3z+2
Step-by-step explanation:
(z+1)(z+1)
Multiply each term in the first parenthesis by each term in the second parenthesis
Z x z+2z+z+2
Calculate the product
<u>z</u>^2 +2z+z+2
collect like terms
z^2+3z+2
2z+z
If a term doesnt have a coefficient it is considered that the coefficient is 1
2z+1z
(2+1)z
(2+1)z
3z
z^2+3z+2
Answer:
15
Step-by-step explanation:
5/15=3/v
5v=15×3
5v=75
5v/5=75/5
v=15
