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2 years ago

What are the constant proportionality y=0.2x

Mathematics

2 answers:

Dmitrij [34]2 years ago

7 0

0.2
y=0.2x
I am pretty sure this is it

Softa [21]2 years ago

6 0

The equation can be written in the form y=kx therefore y varies directly with x and k, the constant of variation, is 0.2.

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Jack and Diane were saving money to go to a rock n roll concert. They need at least $750 in order to go if you count their gas,

nataly862011 [7]

Answer:

$\left\{\begin{array}{l}y\le 4x\\ \\x\ge 10\\ \\50x+25y\ge 750\end{array}\right.$

Step-by-step explanation:

Let x represents the number of nights Jack worked and y represents the number of nights Diane worked.

1. The number of nights Diane is scheduled to work is no more than four times the number of nights Jack is scheduled to play. Then

$y\le 4x$

2. Diane will work at least 10 times before the concert. Then

$x\ge 10$

3. Jack earns $50 per night that he plays, then he earned $50x in x nights. Diane earns $25 each night she works, then she earns $25y in y nigths. They need at least $750, so

$50x+25y\ge 750$

4. We get the following set of constraints to model the problem:

$\left\{\begin{array}{l}y\le 4x\\ \\x\ge 10\\ \\50x+25y\ge 750\end{array}\right.$

4 0

3 years ago

In an automobile parts factory, for every 175 carburetors 3 are found to be defective. At that rate how many carbureatots out of

Furkat [3]

Answer:

At the same rate 18 carburetors out of 1,050 could be expected to be detective.

Step-by-step explanation:

The number of carburetors tested = 175

Defective pieces out o f 175 = 3

So, ratio of defected to tested = 3 : 175

Now, the number of carburetors tested in second batch = 1,050

Here, let the defective pieces = m

So, by the Ratio of Proportion,

$\frac{3}{175} = \frac{m}{1050}$

or, $m = \frac{1050 \times 3}{175} = 18$

⇒ m = 18

Hence, at the same rate 18 carburetors out of 1,050 could be expected to be detective.

8 0

3 years ago

How many kilograms of pasta will each person get if 4 people share 1/2 of a kilogram of pasta equally?

Elenna [48]

Answer:

1/8 kg

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A chicken farmer also has cows for a total of 30 animals, and the animals have 66 legs in all. How many chickens does the farmer

hammer [34]

Answer:

27 chickens

Step-by-step explanation:

A chicken farmer also has cows for a total of 30 animals

Both animals have a total of 66 legs

Let a represent the number of chicken

Let b represent the number of cows

a + b= 30........equation 1

Since a chicken has only two legs and a cow has four leg then, the expression can be represented as

2a + 4b= 66........equation 2

From equation 1

a + b= 30

a = 30-b

Substitute 30-b for a in equation 2

2a + 4b= 66

2(30-b) + 4b= 66

60 - 2b + 4b= 66

60 + 2b= 66

2b= 66-60

2b= 6

b= 6/2

b= 3

Substitute 3 for b in equation 1

a + b = 30

a + 3=30

a= 30-3

a= 27

Hence the farmer has 27 chickens

5 0

3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

case A) $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

case B) $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

case C) $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

case D) $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

2 years ago

Read 2 more answers