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2 years ago

Write a real-world situation that could be modeled by the quation 24/y=3

1 answer:

8 0

There are 24 gallons of water that need to be split up into 3 buckets. How many gallons must each bucket hold? Y=number of gallons that each bucket holds.

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a restaurant served 40 specials in all , 36 of which were the salmon filet . what percentage of the specials were salmon filets?

REY [17]

To find the percentage, you need to divide the 36 by the 40. So 36/40=0.9
0.9=90%

5 0

2 years ago

Jean stocked her aquarium with 36 fresh-water fish, which cost $212. The male fish cost $5 each, while the female fish cost $7 e

alukav5142 [94]

Answer:

Male Fish = 20

Female Fish = 16

Step-by-step explanation:

Let number of male fish be x and number of female fish be y

There are 36 fishes in total, so we can write:

x + y = 36

x = 36 - y

Also, male fish (x) cost 5 each and female fish (y) cost 7 each totalling $212, so we can write:

5x + 7y = 212

Now, we can put 1st equation in 2nd equation and get an equation in "y" and solve for y first:

$5x + 7y = 212\\5(36-y) + 7y = 212\\180-5y+7y=212\\2y=212-180\\2y=32\\y=\frac{32}{2}\\y=16$

We know x = 36 - y, so

x = 36 - 16

x = 20

Thus,

there are

20 Male fishes & 16 female fishes in the aquarium

7 0

2 years ago

Jr tripled the area of his garden by increasing the length by 38. What are the original dimensions

xeze [42]

I confused ! I don't get what your trying to say please explain better !

5 0

2 years ago

What is the value of the radical expression shown below?

avanturin [10]

Answer:

5/7

Step-by-step explanation:

Simplify the radical by breaking the radical up into a product of known factors.

7 0

2 years ago

Read 2 more answers

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