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:
the 2nd
Step-by-step explanation:
Answer:
20 cm
Step-by-step explanation:
So from the problem we get that the two sides of the rctangle are 16 cm 12 cm which are base perpendicular of the right angle triangle. H = 20 cm. So the diagonal of the rectangle = 20 cm.
The Number is 79500, since the nearest thousand for 79500 is 80000
Cº b<span>. </span>Points<span> on the </span>x<span>-axis ( </span>Y. 0)-7<span> (6 </span>2C<span>) are mapped to </span>points<span>. --IN- on the </span>y<span>-axis. ... </span>Describe<span> the transformation: 'Reflect A ALT if A(-5,-1), L(-</span>3,-2), T(-3,2<span>) by the </span>rule<span> (</span>x<span>, </span>y) → (x<span> + </span>3<span>, </span>y<span> + </span>2<span>), then reflect over the </span>y-axis, (x,-1) → (−x,−y<span>). A </span>C-2. L (<span>0.0 tº CD + ... </span>translation<span> of (</span>x,y) → (x–4,y-3)? and moves from (3,-6) to (6,3<span>), by how.</span>
