-9 -4x = 7
add 9 to both sides in order to isolate the x
-4x = 16
divide by -4
x = -4
If we need sections that are 18 ft. long from available 78 ft. long line than we will divide 78/18 =4.(3)
So we will get 4 sections that are 18 ft. long and we will have a remainder of 6 ft line not used.
Answer:
(-2.4, 37.014)
Step-by-step explanation:
We are not told how to approach this problem.
One way would be to graph f(x) = x^5 − 10x^3 + 9x on [-3,3] and then to estimate the max and min of this function on this interval visually. A good graph done on a graphing calculator would be sufficient info for this estimation. My graph, on my TI83 calculator, shows that the relative minimum value of f(x) on this interval is between x=2 and x=3 and is approx. -37; the relative maximum value is between x= -3 and x = -2 and is approx. +37.
Thus, we choose Answer A as closest approx. values of the min and max points on [-3,3]. In Answer A, the max is at (-2.4, 37.014) and the min at (2.4, -37.014.
Optional: Another approach would be to use calculus: we'd differentiate f(x) = x^5 − 10x^3 + 9x, set the resulting derivative = to 0 and solve the resulting equation for x. There would be four x-values, which we'd call "critical values."
Answer:
Step-by-step explanation:
just simplify the LHS first.
You can either multiply 1/5 by (x+3) and then solve
or
multiply both sides by 5 to get rid of 1/5 on LHS
I will multiply by 5
(x+3)= -10x-15 ( 5*1/5(x+3)= -5(2x+3)
now rearrange the equation
x+3=-10x-15
-10x-15-x-3=0
-11x-17=0
-11x=17
x= -17/11
Answer:0.614
Step-by-step explanation:
Given
Probability he will miss flight if it rains=0.06
Probability he will miss flight it does not rain=0.01
Given the probability of rain=0.21
Therefore Probability that it will not rain=1-0.21=0.79
Probability that he will miss the flight 
P(actual raining and he missed the flight| he miss the flight)
