Answer:
There are many solution for this problem
The solutions are -2, -3, -4..... And even fractions like -11/7, -12/7, etc.
Hope this helps!
The diagonal of the tv screen is given by:
d = root ((w) ^ 2 + (h) ^ 2)
Where,
w: width
h: height
In addition we have the following relationship:
h / w = 8/15
Thus,
For the width:
d = root ((w) ^ 2 + ((8/15) w) ^ 2)
For the high:
d = root (((15/8) h) ^ 2 + (h) ^ 2)
Answer:
The expressions for the width and height of a wide-screen tv in terms of the length of its diagonal are:
d = root ((w) ^ 2 + ((8/15) w) ^ 2)
d = root (((15/8) h) ^ 2 + (h) ^ 2)
That’s ur answer, your welcome
I’m pretty sure that it’s b
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
