Answer:
nothing
Step-by-step explanation:
nothing
I will give you everything I can do:
11)
Lets say Car A travels at x mph. That means Car B travels at x+2 mph.
Both of them are traveling towards each others, so we can say the total speed is 2x+2.
Now i takes 3 hrs and we know the distance.
Since R*T=D
Then 3(2x+2)=270
So 2x+2=90
2x=88
x=44
12)
To find perpendicular we want to find the opposite reciprocal of the original slope. Therefore the slope is 3/2.
Now we must find the equation of the line with the given variable.
First find b.
5=3/2*4+b
b = -1
So the equation of this line is:
y=3/2x-1
13) All work will be shown below.
6-3(-2-4x)=2(3(x-4)+7)
6+6+12x=2(3x-12+7)
12+12x=2(3x-5)
12+12x=6x-10
6x=-2
x = -1/3
14)
First we must find the amount each train traveled.
The speed of F train(Freight train)=x
The speed of P train(passenger train)=x+6
Their combined speed is 2x+6
It takes 2 hrs to cover 100 miles
So 2(2x+6)=100
2x+6=50
2x=44
x=22
So the freight train covered 44 miles and the passenger train covered 56 miles.
To find average speed you must do Total Distance/Total Time.
44/2 and 56/2
Which are 22 and 28.
The average speed of F train is 22 mph and average speed of P train is 28 mph.
15) Again opposite reciprocal.
3/5 -> -5/3
Work:
-4=-3*-5/3+b
-4=5+b
b=-9
y = -5/3x-9
16)
F=kx-kx0
First kx0 = 0
So F=kx
So x=F/k
<h3>
Answer: -10 and -40</h3>
===============================================================
Explanation:
a = 200 = first term
d = -30 = common difference
Tn = nth term
Tn = a + d(n-1)
Tn = 200 + (-30)(n-1)
Tn = 200 - 30n + 30
Tn = -30n + 230
Set Tn less than 0 and isolate n
Tn < 0
-30n + 230 < 0
230 < 30n
30n > 230
n > 230/30
n > 7.667 approximately
Rounding up to the nearest whole number gets us 
So Tn starts to turn negative when n = 8
We can see that,
Tn = -30n + 230
T7 = -30*7 + 230
T7 = 20
and
Tn = -30n + 230
T8 = -30*8 + 230
T8 = -10 is the 8th term
and lastly
Tn = -30n + 230
T9 = -30*9 + 230
T9 = -40 is the ninth term
Or once you determine that T7 = 20, you subtract 30 from it to get 20-30 = -10 which is the value of T8. Then T9 = -40 because -10-30 = -40.
The intervals are given as follows:
- In range notation: [-282, 20,320].
- In set-builder notation: {x|x ∈ ℝ, -282 <= x <= 20,320}
<h3>What is the range of elements notation for interval?</h3>
The range of elements notation for interval is given by:
[a,b].
In which:
In this problem these values are given by:
a = -282, b = 20,320.
Hence the interval in range notation is given by:
[-282, 20,320].
<h3>How to write the interval in set-builder notation?</h3>
The same interval can be written as follows, using set-builder notation?
{x|x ∈ ℝ, a <= x <= b}
Hence, for the situation described in this problem, the set-builder notation for the values is:
{x|x ∈ ℝ, -282 <= x <= 20,320}
More can be learned about notation of intervals at brainly.com/question/27896097
#SPJ1
Answer:
The area of ΔABC= 6.667 square units
Step-by-step explanation:
ΔABC is similar to ΔMNO.
The scale factor from ΔMNO to ΔABC is 3∕2
the area of ΔMNO is 10 square units,
The area of ΔABC/the area of ΔMNO
= 2/3
The area of ΔABC/10= 2/3
The area of ΔABC= 2/3 * 10
The area of ΔABC= 20/3
The area of ΔABC= 6 2/3
The area of ΔABC= 6.667 square units
