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2 years ago

How to solve using substitution

Mathematics

1 answer:

DiKsa [7]2 years ago

4 0

X+5 = x^2 - x + 2
0 = x^2 -2x - 3
0 = (x-3)(x+1)

x = -1,3

y = -1 + 5 OR y = 3+5
y = 4 OR y = 8

y = 4,8

So we have two ordered pairs, (-1,4) and (3,8).

Hope that helped!

can you mark my answer as brainliest btw?

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Need help on 15 and 16 PLEASEE!!!

melisa1 [442]

Answer:

Part 15) The next three terms are 5/8,3/4 and 7/8

Part 16) The 37th term is -35.5

Step-by-step explanation:

we know that

In an Arithmetic Sequence the difference between one term and the next is a constant and this constant is called the common difference (d)

Part 15) Find the next three terms of the arithmetic sequence

$\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},...$

$a1=1/8\\ a2=1/4\\ a3=3/8\\ a4=1/2\\ a2-a1=1/4-1/8=1/8\\ a3-a2=3/8-1/4=1/8\\ a4=a3=1/2-3/8=1/8$

The common difference is equal to

d=1/8

Find the next three terms of the arithmetic sequence

Find a5

$a5=a4+d$ -----> $a5=1/2+1/8=5/8$

Find a6

$a6=a5+d$ -----> $a6=5/8+1/8=3/4$

Find a7

$a7=a6+d$ -----> $a7=3/4+1/8=7/8$

therefore

The next three terms are

5/8,3/4 and 7/8

Part 16) what is the 37th term of the arithmetic sequence

$4.1,3,1.9,0.8,...?$

$a1=4.1\\ a2=3\\ a3=1.9\\ a4=0.8$

$a2-a1=3-4.1=-1.1\\ a3-a2=1.9-3=-1.1\\ a4=a3=0.8-1.9=-1.1$

The common difference is equal to

d=-1.1

We can write an Arithmetic Sequence as a rule

$an=a1+d(n-1)$

substitute the values

$an=4.1-1.1(n-1)$

For n=37

substitute

$a37=4.1-1.1(37-1)$

$a37=4.1-1.1(36)$

$a37=-35.5$

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