<span>Plugging in p = 10 to solve for x:
x^2-50x-600 = (x-60)(x+10) = 0 => x = 60
x^2 - 5xp + 50p^2 = 3100
Differentiate with respect to time,
2xx' - 50x'p - 50xp' + 100pp' = 0
240 - 100(10) - 3000p' + 1000p' = 0
p' = -.38
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Answer:
A. 1/5k - 2/3j and -2/3j +1/5k
Step-by-step explanation:
A. 1/5k - 2/3j and -2/3j +1/5k
B. 1/5k - 2/3j and -1/5k +2/3j
There is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
Not equivalent
C. 1/5k - 2/3j and 1/5j - 2/3k
There is a change in the variables
1/5k changed to 1/5j
-2/3j changed to -2/3k
D. 1/5k - 2/3j and 2/3j - 1/5k
The is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
The only equivalent expression is
A. 1/5k - 2/3j and -2/3j +1/5k
A quadratic function is a function of the form

. The
vertex,

of a quadratic function is determined by the formula:

and

; where

is the
x-coordinate of the vertex and

is the
y-coordinate of the vertex. The value of

determines if the <span>
parabola opens upward or downward; if</span>

is positive, the parabola<span> opens upward and the vertex is the
minimum value, but if </span>

is negative <span>the graph opens downward and the vertex is the
maximum value. Since the quadratic function only has one vertex, it </span><span>could not contain both a minimum vertex and a maximum vertex at the same time.</span>
Answer:
Inverse variation k=4
Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent an inverse variation if it can be expressed in the form
or 
In this problem
The graph represent a inverse variation
Because
if x increases ----> y decreases
if x decreases ----> y increases
Find the value of k
we have that
For x=2, y=2 -----> see the graph
substitute
% of 15 in 500= (15÷500)×100
= 0.03×100
= 3 %