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Elenna [48]
3 years ago
7

Y= x +2 and y=3x-4 solved using substitution

Mathematics
1 answer:
weqwewe [10]3 years ago
6 0
So I did this:

y=x +2
y=3x-4

I used the first equation to plug into the second equation.
x+2= 3x -4
I subtracted 3 from both sides so I get
-2x+2=-4
So I subtract 2 from both sides so I get
-2x=-6
I divide -2 on both sides and so I get
x=3
Then I plug the x into either equation. I did the first equation.
y=3+2
y=5
So x=3 and y=5
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Step-by-step explanation:

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Hi guys can u also help me with this, i don’t understand a single thing of it
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HCF=Highest Common Factor-HCF of two or more numbers is the greatest factor that divides the numbers. For example, 2 is the HCF of 4 and 6.
LCM=Least Common Multiple-LCM is the smallest positive number that is a multiple of two or more numbers.

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2 years ago
Simplify 2x(6x^3-4x+11)
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<span><span>12 x^{4} -8 x^{2} +2 2^{x}</span><span>

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8 0
3 years ago
4+2(7)= <br><br> 62+8×3= <br><br> 3(2+5)−5(3)+8= <br><br> 24−6+12÷2×3=
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3(2+5)-5(3)+8= 14

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3 0
4 years ago
A round trip from Montreal to LA is 2,688 miles and takes an airline 9 hours and 20 minutes. Given the outbound trip is the same
Ksju [112]

Answer:

4.074 hours

Step-by-step explanation:

r =

The formula for speed = Distance × Time

Hence,

Let the outbound speed = r

return trip speed = r + 25x/100 = 125r/100 = 1.25r

Note that: 9 hours and 20 minutes = 9.167

Let t1 and t2 be the time taken to travel on way and return journey.

t1 + t2 = 9 hr 10 min = 9.167 hrs

Hence:

2688/r + 2688/1.25r = 9.167

LCM(1.25r)

(2688 × 1.25) + 2688 /1.25r= 9.167

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6048 = 11.45875r

r = 6048/11.45875

Outbound trip rl = 527.8062615904876mph

Return trip speed = r + 25x/100 = 125r/100 = 1.25r

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659.75782699mph

Time for return trip = Distance/Speed

= 2,688 miles/659.75782699mph/

= 4.0742222222 hours

Approximately = 4.074 hours

8 0
3 years ago
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