Answer:
Step-by-step explanation:
From the given information,
Suppose
X represents the Desktop computer
Y represents the DVD Player
Z represents the Two Cars
Given that:
n(X)=275
n(Y)=455
n(Z)=405
n(XUY)=145
n(YUZ)=195
n(XUZ)=110
n((XUYUZ))=265
n(X ∩ Y ∩ Z) = 1000-265
n(X ∩ Y ∩ Z) = 735
n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)
145 = 275+455 - n(X ∩ Y)
n(X ∩ Y) = 585
n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)
195 = 455+405-n(Y ∩ Z)
n(Y ∩ Z) = 665
n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)
110 = 275+405-n(X ∩ Z)
n(X ∩ Z) = 570
a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)
n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735
n(X ∪ Y ∪ Z) = 50
c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)
n(X ∪ Y ∪ C') = 145-50
n(X ∪ Y ∪ C') = 95
Answer:
the nswer is 11.6785714286
Step-by-step explanation:
<span>1) if 2 times the wind speed is increased by 2, the wind speed is still less
than 46 km/h.
=> 2x + 2 < 46
2) Twice the wind speed minus 27 is greater than 11 km/h.
=> 2x - 27 > 11
Part A: Create a compound inequality to represent the wind speed range.
(3 points)
from 2x + 2 < 46
=> 2x < 44
=> x < 22
from 2x - 27 > 11
=> 2x > 11 + 27
=> 2x > 38
=> x > 19
The set of inequalities is
2x + 2 <46
2x - 27 > 11
The solution is x < 22 and x > 19, which is:
19 < x < 22 <----- answer
Part B: Can the wind speed in this town be 20 km/h? Justify
your answer by solving the inequalities in Part A. (3 points)
Yes, the wind speed can be 20 km/h, because the solution of the inequality is the range (19,22).
Part C:
The average wind speed in another town is 23 km/h, but the actual wind
speed is within 4 km/h of the average. Write and solve an inequality to
find the range of wind speed in this town.
x ≥ 23 - 4 => x ≥ 19
x ≤ 23 + 4=> x ≤ 27
=> 19 ≤ x ≤ 27
=> [19,27]
</span>
-(x+2)^2(x-4)^2
Not that if x= -2 and x=4 they will look like this in an equation: (x+2) and (x-4)
When (x+2) and (x-4) are set equal to zero and you solve for x, x will equal x= -2 and x=4
If they have double roots, they have a multiplicity of 2 (per root) meaning they will bounce off of the x-axis. Multiplicity can be found by using exponents.
The negative in front flips the function over the x-axis and holds true to the given limit.
1.
no, there will never be a negative y-value. <span>y= |x| will always be nonnegative. |x| can be distance x is from 0 and a distance can never be negative.
</span>2.
you can define it as
y = |x| = x if x ≥ 0, -x if x < 0
absolute value can be
interpreted as a function that does not allow negative real numbers,
forcing them to be positive (leaving 0 alone). if the input x is more
than or equal 0, then x stays positive so there is no need to do
anything: "x if x ≥ 0".
if the input is less than 0, then it is an
negative number and needs a negative coefficient to negate the negative:
"-x if x < 0"
example: if x = -3, then it will take the "-x if x < 0" piece resulting in y = -(-3) = 3, which is what |-3| does
if x = 1, it will take the "x if x ≥ 0" piece and just have y = 1 which is what |1| does.
for x = 0, it will take the "x if x ≥ 0" and just have y = 0 which is what |0| does