Answer:
C) 0 ≤ x ≤ 25
Step-by-step explanation:
We are supposed to find a reasonable constraint so that the function is at least 300 i.e. the value of x at which f(x) is greater or equal to 300
A)x ≥ 0
Refer the graph
At x = 0
f(x)=300
On increasing the value of x , f(x) increases but at x = 12 it starts decreasing
So, x ≥ 0 can also have f(x)<300
So, Option A is wrong
B)−5 ≤ x ≤ 30
At x = -5
f(x) = 100
So, Option B is wrong since we require f(x) is greater or equal to 300
c)0 ≤ x ≤ 25
At x = 0
f(x)=300
At x = 12 , it starts decreasing
At x = 25
f(x)=300
So, The value of f(x) is at least 300 when 0 ≤ x ≤ 25
D)All real numbers
At x = 30
f(x)=0
But we require f(x) greater or equal to 300
Hence Option C is true
Answer: x=-1 y=1
Step-by-step explanation:
solve y in 3+2x-y=0
y=3+2x
sub y=3+2x into -3 -7y=10x
-14x- 24= 10x
solve x in -14x-24=10x
x= -1
sub x= -1 into y =3+2x
y=1
so therefore x= -1 and y=1
15. 3x - 2y = -2
3x - 3x - 2y = -3x - 2
-2y = -3x - 2
-2 -2
y = 1.5x + 1
y - y₁ = m(x - x₁)
y - 3 = ⁻²/₃(x - (-2))
y - 3 = ⁻²/₃(x + 2)
y - 3 = ²/₃(x) - ²/₃(2)
y - 3 = ⁻²/₃x - 1¹/₃
+ 3 + 3
y = ⁻²/₃x + 1²/₃
16. 230 = 0.2s + 150
- 150 - 150
80 = 0.2s
0.2 0.2
400 = s
17. y = 2x + 2
