What is 1/5 divided by 1 2/5

Put 5x – 6y = 12 in slope intercept form

Ede4ka [16]

Answer:

y = 5/6x -2

Step-by-step explanation:

Subtract 5x and divide by -6:

-6y = -5x +12 . . . .subtract 5x

y = 5/6x -2 . . . . . divide by -6

6 0

4 years ago

If the present value of a growing perpetuity is 214, the required rate of return is 10%, and growth rate is 3%, what is the cash

Kryger [21]

The cash flow in year 1 is $15.

A growing perpetuity increases continuously and indefinitely.

Cash flow = present value x (rate of return - growth rate)

214 x (10% - 3%)

214 x 0.07 = $15

To learn more about present value, please check: brainly.com/question/17322936

#SPJ1

3 0

2 years ago

1. What is the solution of the system?

andreyandreev [35.5K]

Answer:

The given system of equations has solutions below:

1) The solution is (2,3)

2) The solution is ( $\frac{-8}{7}, \frac{5}{7}$ )

3) The solution is infinitely many solutions

4) No solution

Step-by-step explanation:

Given system of equation are

$-x+2y=4\hfill (1)$

$-4x+y=-5\hfill (2)$

To solve equation by using elimination method

Multiply eqn (2) into 2

$-8x+2y=-10\hfill (3)$

Now subtracting (1) and (3)

$-x+2y=4$

$-8x+2y=-10$

_________________

7x=14

x= $\frac{14}{7}$

$x=2$

Substitute x=2 in equation (1)

-2+2y=4

2y=4+2

$y=\frac{6}{2}$

y=3

Therefore the solution is (2,3)

2) Given equation is

$-2x+y=3\hfill (1)$

4y-4=x

Rewritting as below

$x-4y=-4\hfill (2)$

To solve equation by using elimination method

multiply (2) into 2

$2x-8y=-8\hfill (3)$

Adding (1) and (3)

-2x+y=3

2x-8y=-8

________

-7y=-5

$y=\frac{5}{7}$

substitute $y=\frac{5}{7}$ in (1)

$-2x+\frac{5}{7}=3$

$-2x=3-\frac{5}{7}$

$-2x=\frac{21-5}{7}$

$x=-\frac{8}{7}$

Therefore the solution is ( $\frac{-8}{7},\frac{5}{7}$ )

3) Given equation is $6x+2y=10\hfill (1)$

$3x+y=5\hfill (2)$

equation (1) can be written as

2(3x+y)=10

$3x+y=\frac{10}{2}$

3x+y=5

Therefore equations (1) and (2) are same therefore it has infinitely many solutions

4) Given equation is $-x-2y=14\hfill (1)$

$-2x-4y=12\hfill (2)$

multiply equation (1) into 2

$-2x-4y=28\hfill (3)$

To solve equation by using elimination method

subtracting (2) and (3)

-2x-4y=28

-2x-4y=12

_______

$28\neq -12$

therefore it has no solution

4 0

3 years ago

Helppppppppppp:)))))))))

Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

$\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}$

1. What is the domain?

Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.

Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, the actual domain is {0,2,3}

2. What is the range?

Because domain is set of all x-values. Then what do you think the range is? That's right! The range is set of all y-values. If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!

Therefore, gather y-values from the ordered pairs. Hence, the range is {-2,-1,1,2}

3. Is the relation a function?

All functions are relations but not all relations are functions. Function is a set of ordered pairs where domain is not repetitive or in a set, there cannot be more than one same value. Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.

Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.

These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.

From the Q1's answer, there are two bold texts, please choose the second bold text to answer (the one with underline) and not the first one (the one with same 2's).

Good luck on your assignment, have a nice day!

4 0

3 years ago

PLEASE HELPPP!!! I will mark brainlist

Maslowich

Answer:

d

Step-by-step explanation:

5 0

3 years ago

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