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crimeas [40]
3 years ago
12

Find each percent decrease. round to the nearest percent. from $16 to $4​

Mathematics
1 answer:
White raven [17]3 years ago
8 0

Answer:

I think the answer would be 40%

Step-by-step explanation:

U would do 16 divided by 4 and get 4 and a 0 then bam I have ur answe.

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WILL MARK BRAINLIEST
777dan777 [17]

Answer:


Step-by-step explanation:

xy = k    

where k is the constant of variation.

We can also express the relationship between x and y as:

y =      

where k is the constant of variation.

Since k is constant, we can find k given any point by multiplying the x-coordinate by the y-coordinate. For example, if y varies inversely as x, and x = 5 when y = 2, then the constant of variation is k = xy = 5(2) = 10. Thus, the equation describing this inverse variation is xy = 10 or y = .

Example 1: If y varies inversely as x, and y = 6 when x = , write an equation describing this inverse variation.

k = (6) = 8  

xy = 8 or y =  

Example 2: If y varies inversely as x, and the constant of variation is k = , what is y when x = 10?

xy =  

10y =  

y = × = × =  

k is constant. Thus, given any two points (x1, y1) and (x2, y2) which satisfy the inverse variation, x1y1 = k and x2y2 = k. Consequently, x1y1 = x2y2 for any two points that satisfy the inverse variation.

Example 3: If y varies inversely as x, and y = 10 when x = 6, then what is y when x = 15?

x1y1 = x2y2  

6(10) = 15y  

60 = 15y  

y = 4  

Thus, when x = 6, y = 4.



2nd answer choice


constant of variation is xy. XY=23. If X=7 then Y=23/7.

3 0
3 years ago
Add using a number line.<br> +(-11)
Nikolay [14]

Answer:

-11 is the anwer.

As it is present in left side, so -11 will be in left part of number line.

4 0
3 years ago
How do you solve this?
oksian1 [2.3K]

Answer:

X=73/7

Step-by-step explanation:

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3 years ago
Need help with functions
Komok [63]

Answer: ur right

Step-by-step explanation:

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2 years ago
Which equation can you use to solve for figure shown?
DIA [1.3K]

Hey mate

Here is ur answer ...

I hope it'll help ....

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3 years ago
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