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Korolek [52]
4 years ago
10

Solve the simultaneous equations y=x-2 and y=3x+5

Mathematics
1 answer:
IgorLugansk [536]4 years ago
3 0

Answer:

(- 3.5, - 5.5 )

Step-by-step explanation:

Given the 2 equations

y = x - 2 → (1)

y = 3x + 5 → (2)

Substitute y = 3x + 5 into (1)

3x + 5 = x - 2 ( subtract x from both sides )

2x + 5 = - 2 ( subtract 5 from both sides )

2x = - 7 ( divide both sides by 2 )

x = - 3.5

Substitute x = - 3.5 into either of the 2 equations and evaluate for y

Substituting into (1)

y = - 3.5 - 2 = - 5.5

Solution is (- 3.5, - 5.5 )

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Find the sum of the first 8 terms in the sequence 2 8 32 128
Zigmanuir [339]

the sequence is in multiples of 4 (2*4=8, 8*4=32, 32*4=128)

so 128*4=512, 512*4=2048, 2048*4=8192 & 8192*4=32768

2+8+32+128+512+2048+8192+32768=43690 

4 0
3 years ago
What must be added to -16 to give 45​
german
The answer is: positive 61.
7 0
3 years ago
Pierce works at a tutoring center on the weekends. At the center, they have a large calculator to use for demonstration purposes
icang [17]

Answer:

504 millimeters (or 50.4 cm)

Step-by-step explanation:

   Width of key in student calculator = 14 millimeter (1.4 cm)

   Width of key in demonstration calculator = 2.8 cm

Thus, the demonstration calculator's dimensions are twice that of students' (in cm)

Also given, student calculator height as 252 millimeters (25.2 cm)

Thus demonstration calculator height will be twice of that = 50.4 cm (or 504 millimeters)

5 0
3 years ago
What is the product of -3(-6r+6)
Vladimir [108]

Step-by-step explanation:

-3 times -6r= 18r

-3 times 6 =-18

18r=-18

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If the answer is wrong try just 1

8 0
4 years ago
Read 2 more answers
24% of 1,321 is how much?
cupoosta [38]
Percentage that needs to be determined = 24%
The quantity from which the percentage has to be determined = 1321
Then
24% of 1321 = (24/100) * 1321
                     = 31704/100
                     = 317.04
So from the above deduction we can easily say that 24% of the number 1321 is 317.04. I hope the procedure is not very complicated for you to understand. You can use this method for solving similar type of problems in future without requiring any help from anyone. Only be careful while calculating. Other than the calculation part, there is nothing tough about the problem.
5 0
3 years ago
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