Answer:
А.The system has two solutions, but only one is viable because the other results in a negative width.
Step-by-step explanation:
Given
Let:
length of play area A
width of play area A
length of play area B
width of play area B
Area of A
Area of B
From the question, we have the following:
The area of A is:
This gives:
Open bracket
The area of B is:
Substitute:
Open brackets
Expand
We have that:
This gives:
Collect like terms
Using quadratic calculator, we have:
or 
--- approximated
But the width can not be negative; So:
Answer:
26: x=34
28: x=17
30: m=30
Step-by-step explanation:
<u>26:</u> 2x+22=90
subtract 22
2x=68
divide by 2
x=34
<u>28:</u> 18+3x+21=90
add 18 and 21
3x+39=90
subtract 39
3x=51
divide by 3
x=17
<u>30:</u> 43+87+m+20=180
add 43 and 87 and 20
m+150=180
subtract 150
m=30
Combine like terms.
4x^3 -6x^3= -2x^3
-8x^2 +2x^2= -6x^2
Answer = -2x^3 -6x^2 +4
Answer:
f(7) = 158
Step-by-step explanation:
To evaluate f(7) substitute t = 7 into f(t)
f(7) = 3(7)² + 11 = (3 × 49) + 11 = 147 + 11 = 158
