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3 years ago

Y=(3/5)x -3 and 5y =3x-10 these lines are and the second question?

Mathematics

1 answer:

azamat3 years ago

4 0

Look at the first line: y = (3/5)x - 3. What happens if you multiply each term by 5, to eliminate the fraction?

5y = 3x - 3

Compare this to the second equation,

5y - 3x = -10, or 5y = 3x - 10.

The coefficients of x and y (as 3 and 5 here) determine the slope of a straight line. Since 5y = 3x is present in both equations, the two lines MUST be parallel.

y = 4
4y = 6 => y = 6/4

y+4 and y =3/2 are both horizontal lines. Since they are horiz., they are parallel.

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A sequence is defined by the formula f(n + 1) = f(n) – 3. If f(4) = 22, what is f(1)?

julsineya [31]

f(n) = f(n+1) +3

f(1) = f(2) +3

f(2) =f(3) +3

f(3) = f(4) +3

So f(1) = f(n) + 3* (n-1)

Put n=4 here so you found

f(1) = f(4) + 3* 3

= 22 +9

=31 ans

hope it helps

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Answer:

$\huge\boxed{14\ \text{and}\ 7}$

Step-by-step explanation:

$n,\ m-\text{positive integer}\\\\n=2m-\text{a positive integer is twice another}\\\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}-\text{the sum of the reciprocal of the two positive integer is }\ \dfrac{3}{14}\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}n=2m&(1)\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}&(2)\end{array}\right$

$\text{Substitute (1) to (2):}\\\\\dfrac{1}{2m}+\dfrac{1}{m}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{1\cdot2}{m\cdot2}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{2}{2m}=\dfrac{3}{14}\\\\\dfrac{1+2}{2m}=\dfrac{3}{14}\\\\\dfrac{3}{2m}=\dfrac{3}{14}\Rightarrow2m=14\qquad\text{divide both sides by 2}\\\\\dfrac{2m}{2}=\dfrac{14}{2}\\\\\boxed{m=7}$

$\text{Substitute it to (1):}\\\\n=2\cdot7\\\\\boxed{n=14}$

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