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NikAS [45]
3 years ago
9

Answer two questions about Systems AAA and BBB: System AAA \text{\quad}start text, end text System BBB \begin{cases}x-7y=-3\\\\4

x+y=7\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ x−7y=−3 4x+y=7 ​ \begin{cases}4x+y=7\\\\x-7y=-3\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ 4x+y=7 x−7y=−3 ​ 1) How can we get System BBB from System AAA?
Mathematics
2 answers:
MrRa [10]3 years ago
5 0

Answer:

B: swap the order of the equations

(Yes) they have the same solution

Step-by-step explanation:

GenaCL600 [577]3 years ago
5 0

Answer:

B

Yes

Step-by-step explanation:

Hope this helps!

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Graph B

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Can the average rate of change of a function be constant ? Explain
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  yes

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The average rate of change of any linear function is a constant. That fact is what makes it a linear function.

8 0
3 years ago
Can somebody please help me with this question?
Aliun [14]

Answer:

120, 114, 126

Step-by-step explanation:

Since these are all exterior angles, they will all always add up to 360°. Since we know this, you can set up an equation and it would be x+(x+6)+114=360. When added you get 2x+120=360. Now subtract 120 on each side to get 2x equals 240. Then you divide by 2 and get that x is 120. Now all you have to do is plug x into the equations for the angles so they would be 120, 114, and 120+6 so 126. Now you have 120,114,and 126 as your angles. To check, just add all the angle measurements up and it should equal 360.

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HELP PLEASE CAN’T FIGURE THIS OUT NEED HELP
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Step-by-step explanation:

6 0
3 years ago
Car A, traveling at a constant 10 m/s, crosses the start line 3 seconds before Car B, who is traveling at a constant 15 m/s. At
Margaret [11]
The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
 
Distance traveled by car A =x,  (when the car B is at the same distance from the start line)
time of car A=t

x=10 m/st  ⇒    x=10t    (1)

Distance traveled by car B=x
time of car B=t-3

x=15(t-3)  (2)

With the equations (1) and (2) we make a system of equations:

x=10t
x=15(t-3)

We solve this system of equations:

10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9

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x=10 t=10 (9)=90

Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.
8 0
3 years ago
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