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2 years ago

Find k, the constant of proportionality, for the data in this table. Then write an equation for the relationship.

Mathematics

1 answer:

VashaNatasha [74]2 years ago

7 0

Answer:

$k=\frac{32}{5}$ , $y=\frac{32}{5}x$

$k=6.4$ , $y=6.4x$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

Find the value of the constant of proportionality k

take any ordered pair from the data

For x=25, y=160

$k=\frac{y}{x}$

substitute the values of x and y

$k=\frac{160}{25}$

simplify

$k=\frac{32}{5}$

The linear equation is equal to

$y=\frac{32}{5}x$

$y=6.4x$

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Giovani solves a system of equations algebrically he concludes that the lines are parallel which could be the final of his solut

kenny6666 [7]

Answer:

If the lines are parallel, final form of his solution is

b) -5 = -5+1

Step-by-step explanation:

As the options are not given in the question and the question is incomplete, lets complete the question first.

Giovani solves a system of linear equations algebrically, he concludes that the lines are parallel. Which could be his final answer.

a) -5 = -5

b) -5 = -5+1

c) 11 = 11

d) 5 = 5

We know that parallel lines never cross at any point, there is no intersection point, which eventually means that there is no solution to the given equations. A system of equations which have no solution is called an inconsistent system of equations, and it graphs the parallel line.

From the options we can clearly see that only (b) has no solution, while all the rest of the options have solutions. Hence, -5 = -5+1 represents parallel lines.

3 0

2 years ago

Please help. 24 points!!

Lina20 [59]

The given function is,

$f(x)= \frac{2x-1}{x+5}$ .

Or, $y= \frac{2x-1}{x+5}$

To find the inverse of a function, first step is to switch x and y. Therefore,

$x= \frac{2y-1}{y+5}$

Next stpe is to solve the above equation for y t get the inverse f f(x). Hence, multiply each sides by y + 5 to get rid of fraction form. So,

$x*(y+5)= \frac{2y-1}{y+5}*(y+5)$

xy + 5x = 2y - 1

xy = 2y - 1 - 5x Subtract 5x from each sides to isolate y.

xy - 2y = -5x - 1 Subtract 2y from each sides.

y(x - 2) = -5x -1 Take out the common factor y.

$\frac{y(x-2)}{(x-2)} =\frac{(-5x-1)}{(x-2)}$ Divide each sides by x-2.

$y=\frac{(-5x-1)}{(x-2)}$

So, $f^-1(x)=\frac{-5x-1}{x-2}$

Now when we will compare $f^-1(x)=\frac{-5x-1}{x-2}$ with $f^-1(x)=\frac{ax+b}{cx+d}$ then we will get a = -5, b = -1, c = 1 and d=-2

So, $\frac{a}{c} =\frac{-5}{1} =-5$

Hope this helps you!

6 0

2 years ago

Hey guys please help with this questoin Find (x+y) ÷ (x-y) if x=5/4 y=-1/3

pogonyaev

Answer:

Step-by-step explanation:

$(\frac{5}{4} + (-\frac{1}{3})) \div (\frac{5}{4} - (-\frac{1}{3}))\\\\\frac{\frac{5}{4}+\left(-\frac{1}{3}\right)}{\frac{5}{4}-\left(-\frac{1}{3}\right)}\\\\\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a\\=\frac{\frac{5}{4}-\frac{1}{3}}{\frac{5}{4}+\frac{1}{3}}\\\\\mathrm{Join}\:\frac{5}{4}+\frac{1}{3}:\quad \frac{19}{12}\\\mathrm{Join}\:\frac{5}{4}-\frac{1}{3}:\quad \frac{11}{12}\\\\=\frac{\frac{11}{12}}{\frac{19}{12}}\\\\=\frac{11\times\:12}{12\times \:19}$