Answer:
4. Circumference = diameter × π
8× 3. 14= 25. 12 ~25. 1
5. 13×3. 14= 40. 82 ~40. 8
6. Diameter= radius ×2
3. 5×2=7
Diameter = 7
7× 3. 14= 21. 98 ~22
Answer:
I) f + g = 10/3
II) 4f + 2g = 20/3
III) f = 2 and g = 4/3
Step-by-step explanation:
From the chart,
P = 25
q = 40
The total number of one centimeter lines in the first n diagrams is given by the expression
2/3n^3 + fn^2 + gn.
When n = 1, the total number of line = 4. So,
2/3(1)^3 + f(1)^2 + g(1) = 4
2/3 + f + g = 4
Make f+g the subject of formula
f + g = 4 - 2/3
f + g = (12 - 2)/3
f + g = 10/3 ......(1)
When n = 2
Total number of line = 12
2/3(2)^3 + f(2)^2 + g(2) = 12
2/3×8 + 4f + 2g = 12
16/3 + 4f + 2g = 12
4f + 2g = 12 - 16/3
4f + 2g = (36 - 16)/3
4f + 2g = 20/3 ......(2)
(iii) To find the values of f and g, solve equation 1 and 2 simultaneously
f + g = 10/3 × 2
4f + 2g = 32/3
2f + 2g = 20/3
4f + 2g = 32/3
- 2f = - 12/3
f = 12/6
f = 2
Substitutes f in equation 1
f + g = 10/3
2 + g = 10/3
g = 10/3 - 2
g = (10 - 6)/3
g = 4/3
Answer:
Let's define the variables:
A = price of one adult ticket.
S = price of one student ticket.
We know that:
"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."
1*A + 6*S = $69
"The school took in $150 on the second day by selling 7 adult tickets and student tickets"
7*A + 7*S = $150
Then we have a system of equations:
A + 6*S = $69
7*A + 7*S = $150.
To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:
A = $69 - 6*S
Now let's replace this in the other equation:
7*($69 - 6*S) + 7*S = $150
Now we can solve this for S.
$483 - 42*S + 7*S = $150
$483 - 35*S = $150
$483 - $150 = 35*S
$333 = 35*S
$333/35 = S
$9.51 = S
That we could round to $9.50
That is the price of one student ticket.
Answer:
I think you solved it my g