Most of the time its juss common sense where you can point the right measurement with just sense i guess
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
Answer:
k = 2
Z = 3.5
Step-by-step explanation:
Z = kX/Y
Where,
X = 3
Y = 1
Z = 6
Find k
Z = kX/Y
6 = k3/1
Cross product
6 * 1 = 3k
6 = 3k
k = 6/3
k = 2
what is the value of Z when X equals seven and why equals four
X = 7
Y = 4
Z = ?
k = 2
Z = kX/Y
= (2*7)/4
= 14/4
= 3.5
Z = 3.5
The answer is A
y= 0.3x + 2.9
Answer:
11
Step-by-step explanation:
add up all of the X symbols to get 11. Hope i helped!