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Harlamova29_29 [7]
2 years ago
9

There are Z fish in a aquarium. 1/4 of the fish are angelfish. How may are angelfish?

Mathematics
2 answers:
likoan [24]2 years ago
8 0
The answer is 1/4*z=1/4z
blsea [12.9K]2 years ago
7 0

Step-by-step explanation:

since there are z fishes

the number of angelfish = <u>1</u><u> </u> × z

4

= <u>1</u><u> </u>z

4

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How many zeros does the function f(x) = 2x14 − 14x6 27x3 − 13x 12 have?
Anettt [7]

Answer:

14 zeros

Step-by-step explanation:

For functions, the highest power is the amount of zeros. The highest power in this equation is 14 so it has 14 zeros.

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Xelga [282]

Answer:

D. y=162.5

Step-by-step explanation:

0.75 ->25

4.875->y

y= (4.875*25)/0.75

y=162.5

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(6x-5y+4)dy+(y-2x-1)dx=0​
Len [333]

(6<em>x</em> - 5<em>y</em> + 4) d<em>y</em> + (<em>y</em> - 2<em>x</em> - 1) d<em>x</em> = 0

(6<em>x</em> - 5<em>y</em> + 4) d<em>y</em> = (2<em>x</em> - <em>y</em> + 1) d<em>x</em>

d<em>y</em>/d<em>x</em> = (2<em>x</em> - <em>y</em> + 1) / (6<em>x</em> - 5<em>y</em> + 4)

Let <em>X</em> = <em>x</em> - <em>a</em> and <em>Y</em> = <em>y</em> - <em>b</em>. We want to find constants <em>a</em> and <em>b</em> such that

d<em>Y</em>/d<em>X</em> = (a rational function)

where the numerator and denominator on the right side are free of constant terms. Substituting <em>x</em> and <em>y</em> in the equation, we have

d<em>Y</em>/d<em>X</em> = (2 (<em>X</em> + <em>a</em>) - (<em>Y</em> + <em>b</em>) + 1) / (6 (<em>X</em> + <em>a</em>) - 5 (<em>Y</em> + <em>b</em>) + 4)

d<em>Y</em>/d<em>X</em> = (2<em>X</em> - <em>Y</em> + 2<em>a</em> - <em>b</em> + 1) / (6<em>X</em> - 5<em>Y</em> + 6<em>a</em> - 5<em>b</em> + 4)

Then we solve for <em>a</em> and <em>b</em> in the system,

2<em>a</em> - <em>b</em> + 1 = 0

6<em>a</em> - 5<em>b</em> + 4 = 0

==>   <em>a</em> = -1/4 and <em>b</em> = 1/2

With these constants, the equation reduces to

d<em>Y</em>/d<em>X</em> = (2<em>X</em> - <em>Y</em>) / (6<em>X</em> - 5<em>Y</em>)

Now substitute <em>Y</em> = <em>VX</em> and d<em>Y</em>/d<em>X</em> = <em>X</em> d<em>V</em>/d<em>X</em> + <em>V</em> :

<em>X</em> d<em>V</em>/d<em>X</em> + <em>V</em> = (2<em>X</em> - <em>VX</em>) / (6<em>X</em> - 5<em>VX</em>)

The equation becomes separable after some simplification:

<em>X</em> d<em>V</em>/d<em>X</em> + <em>V</em> = (2 - <em>V</em>) / (6 - 5<em>V</em>)

<em>X</em> d<em>V</em>/d<em>X</em> = (2 - <em>V</em>) / (6 - 5<em>V</em>) - <em>V</em>

<em>X</em> d<em>V</em>/d<em>X</em> = (2 - <em>V</em> - (6 - 5<em>V</em>)) / (6 - 5<em>V</em>)

<em>X</em> d<em>V</em>/d<em>X</em> = (4<em>V</em> - 4) / (6 - 5<em>V</em>)

- (5<em>V</em> - 6) / (4<em>V</em> - 4) d<em>V</em> = 1/<em>X</em> d<em>X</em>

Integrate both sides:

-5/4 <em>V</em> + 1/4 ln|4<em>V</em> - 4| = ln|<em>X</em>| + <em>C</em>

Extract a constant from the logarithm on the left:

-5/4 <em>V</em> + 1/4 (ln(4) + ln|<em>V</em> - 1|) = ln|<em>X</em>| + <em>C</em>

-5/4 <em>V</em> + 1/4 ln|<em>V</em> - 1| = ln|<em>X</em>| + <em>C</em>

-5<em>V</em> + ln|<em>V</em> - 1| = 4 ln|<em>X</em>| + <em>C</em>

Get this back in terms of <em>Y</em> :

-5<em>Y/X</em> + ln|<em>Y/X</em> - 1| = 4 ln|<em>X</em>| + <em>C</em>

Now get the solution in terms of <em>y</em> and <em>x</em> :

-5 (<em>y</em> - 1/2)/(<em>x</em> + 1/4) + ln|(<em>y</em> - 1/2)/(<em>x</em> + 1/4) - 1| = 4 ln|<em>x</em> + 1/4| + <em>C</em>

<em />

With some manipulation of constants and logarithms, and a bit of algebra, we can rewrite this solution as

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-5 (4<em>y</em> - 2)/(4<em>x</em> + 1) + ln|(4<em>y</em> - 4<em>x</em> - 3)/(4<em>x</em> + 1)| = 4 ln|4<em>x</em> + 1| + <em>C</em>

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S_A_V [24]
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dangina [55]
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