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sergey [27]
3 years ago
11

4/x - 3/y = 1 ; 6/x + 15/y = 8 solve this equation​

Mathematics
1 answer:
Soloha48 [4]3 years ago
4 0

Answer:

x=2

y=3

Solution:

First we find common denominators. It is "xy". Then we multiply numerators by common denominator. We get followings:

(4y-3x)/xy=1; (6y+15x)/xy=8

Then

4y-3x=xy;

6y+15=8xy

Multiply first equasion by 5

20y-15x=5xy

Now we add two equasions to get one

20y-15x=5xy

6y+15x=8xy

We get

26y=13xy

Cut "y" and we will find "x"

26=13x

x=2

Put x value into the first equasion(4y-3x=xy) to find out "y"

4y-6=2y

2y=6

y=3

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A sample of final exam scores is normally distributed with a mean equal to 23 and a variance equal to 16. Part (a) What percenta
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Answer:

Let X = score of final exam.

X~normal(23, 16)

(a)

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4 years ago
Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B​
telo118 [61]

In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

Let's write both sides in terms of \sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x) only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)

and

\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)

So, their multiplication is

\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)

So, the left hand side simplifies to

\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

And as you can see, the two sides are equal.

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4 years ago
A recent school survey found that 8 out of 10 students plan to go to college after graduating high school. At that rate, how man
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Answer:

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