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3 years ago

Which ordered pair is the solution?

Mathematics

2 answers:

Alex777 [14]3 years ago

8 0

Answer:

Option d.

Step-by-step explanation:

The given equation is

$7x-5=4y-6$

We need check which ordered pair is the solution.

Substitute x=2 and y=4 in the above equation.

$7(2)-5=4(4)-6$

$14-5=16-6$

$9=10$

It is a false statement because $9\neq 10$ .

So, (2,4) is not a solution of given equation.

Substitute x=3 and y=6 in the above equation.

$7(3)-5=4(6)-6$

$21-5=24-6$

$16=18$

It is a false statement because $16\neq 18$ .

So, (3,6) is not a solution of given equation.

Therefore, option d is correct.

Vlad [161]3 years ago

6 0

Answer:

d) neither

Step-by-step explanation:

Given equation is 7x - 5 = 4y -6 ...…...... (1)

Substitute x = 2 and y= 4 in equation (1)

7(2) -5 = 4(4) -6

9 ≠ 10

a) is not satisfied

Substitute x = 3 and y= 6 in equation (1)

7(3) -5 = 4(6) -6

16 ≠ 18

b) is not satisfied

c) is also not satisfied

Final answer:- neither

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Answer:

Part A

$f(n)=52-12(n-1)$

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$ Geometric Sequence

Part C

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Part D:

$h(n)=1.1+0.4(n-1)\\h(n)=\left\{\begin{matrix}1.1 & if\:n=1 \\ h(n+1)+0.4 & if\:n\geq 2\end{matrix}\right$

Step-by-step explanation:

By definition, an Arithmetic Sequence holds the same difference between each following number.

Part A

$(52,40, 28, 16)\\52-40=12\\40-28=12\\28-16=12\\d=12$

Explicit Formula

To write an explicit formula is to write it as function.

$f(n)=52-12(n-1)$

Recursive Formula

To write it as recursive formula, is to write it as recurrence given to some restrictions:

$f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.$

Part B

$(2,4,8,16,32)\: \:$

Geometric Sequence, since 2*2=4 8*2=16 and 16*2=32 and 8+2=10 8+16=24

Part C

$(\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4})\\\$

Arithmetic Sequence, difference

$d=\frac{2}{4}$

Explicit Formula:

$g(n)=\frac{1}{4}+\frac{2}{4}(n-1)$

Recursive Formula

$g(n)=\left\{\begin{matrix}\frac{1}{4} &if\:n=1 \\ g(n+1)+\frac{2}{4} &if\: n\geq 2\end{matrix}\right.$

Part D

(1.1,1.5,1.9,2.3,2.7) Arithmetic Sequence, difference d=0.4

Explicit formula

$h(n)=1.1+0.4(n-1)\\$

Recursive Formula

$h(n)=\left\{\begin{matrix}1.1 &if\:n=1 \\ h(n+1)+0.4 &if\: n\geq 2\end{matrix}\right.$

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