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3 years ago

Please help! show work if you can

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

Step-by-step explanation:

i did test please comment if its wrong

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Please help me with this question thank you

gulaghasi [49]

Answer:

$\mathrm{The\:solution\:is}$ :

$x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

Step-by-step explanation:

Given

$y=\sqrt{\frac{2x+1}{x-1}}$

Taking square of both sides

$y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)\:^2$

$\mathrm{Subtract\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2\mathrm{\:from\:both\:sides}$

$y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Simplify}$

$y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=0$

As we know that $\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2:\quad \left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)$

$\left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$\mathrm{Solve\:}\:y+\sqrt{\frac{2x+1}{x-1}}=0$

$\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides}$

$y+\sqrt{\frac{2x+1}{x-1}}-y=0-y$

$\sqrt{\frac{2x+1}{x-1}}=-y$

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2$

$\mathrm{Expand\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}$

$=\left(\left(\frac{2x+1}{x-1}\right)^{\frac{1}{2}}\right)^2$

$=\frac{2x+1}{x-1}$

so equation $\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2$ becomes

$\frac{2x+1}{x-1}=y^2$

now

$\mathrm{Solve\:}\:\frac{2x+1}{x-1}=y^2$

$\frac{2x+1}{x-1}=y^2$

$\mathrm{Multiply\:both\:sides\:by\:}x-1$

$\frac{2x+1}{x-1}\left(x-1\right)=y^2\left(x-1\right)$

$2x+1=y^2\left(x-1\right)$

$2x+1=xy^2-y^2$ ∵ $y^2\left(x-1\right):\quad xy^2-y^2$

$2x=xy^2-y^2-1$

$2x-xy^2=-y^2-1$

$x\left(2-y^2\right)=-y^2-1$ ∵ $\mathrm{Factor}\:2x-xy^2:\quad x\left(2-y^2\right)$

$\mathrm{Divide\:both\:sides\:by\:}2-y^2$

$\frac{x\left(2-y^2\right)}{2-y^2}=-\frac{y^2}{2-y^2}-\frac{1}{2-y^2}$

$x=\frac{-y^2-1}{2-y^2}$

$y+\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\le \:0\right\}$

similarly

$y-\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

$\mathrm{Verify\:Solutions}:\quad x=\frac{-y^2-1}{2-y^2}$

$\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2$

$\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}$

$\mathrm{Plug}\quad x=\frac{-y^2-1}{2-y^2}$

$y^2=\left(\sqrt{\frac{2\left(\frac{-y^2-1}{2-y^2}\right)+1}{\left(\frac{-y^2-1}{2-y^2}\right)-1}}\right)^2$

$\mathrm{Subtract\:}\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2\mathrm{\:from\:both\:sides}$

$y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2$

$y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=0$

$\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2:\quad \left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)$

$\left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$\mathrm{Solve\:}\:y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\le \:0$

$\mathrm{Solve\:}\:y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\ge \:0$

$\mathrm{True\:for\:all}\:y$

Therefore, $\mathrm{The\:solution\:is}$ :

$x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}$

6 0

3 years ago

2 is what percent of 10

kolezko [41]

Answer:

2 is 20% of 10

Step-by-step explanation:

hope this helps

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4 0

2 years ago

given three points of a quadratic function find the equation that defines the function (-2,20)(0,-2)(1,-4) hurry please!!

kotykmax [81]

Answer:

D: y = 3x² - 5x - 2

Step-by-step explanation:

The general form of a quadratic equation is: y = ax² + bx + c

c is the y-intercept

We are given the point (0, -2) which is the y-intercept, so we can rewrite our general form into

y = ax² + bx - 2

We can create a system of equations to solve for a and b. We are given two points.

Equation 1: Take the first point (-2, 20) and plug it into our general equation...

20 = a(-2)² + b(-2) - 2

20 = 4a - 2b - 2

22 = 4a - 2b (add 2 to both sides)

11 = 2a - b (divide both sides by 2 since every coefficient is even)

Equation 2: Take the point (1, -4) and plug it into the general equation

-4 = a(1)² + b(1) - 2

-4 = a + b - 2

-2 = a + b

Now we have our 2 equations:

11 = 2a - b

-2 = a + b

Since the coefficients of b are already have opposite signs, add the two equations together (elimination method)

Now we have

9 = 3a now solve for a...

3 = a (divide by 3 on both sides)

If a = 3, then

-2 = 3 + b

-5 = b

Our equation is

y = 3x² - 5x - 2

6 0

3 years ago

Find the y-coordinate of the y-intercept of the polynomial function defined

Bas_tet [7]

The y intercept is where the function passes through the y axis, so the “x” value needs to equal 0. We sub 0 in for “x” and solve for “y”.

f(0) = 9(0)^2 - 7 - 8(0)

f(0) = -7

Therefore the “y” coordinate would be -7

The ordered pair would be

(0, -7)

6 0

3 years ago

An animal shelter spends $2.35 per day to care for each cat and $5.50 per day to care for each

lora16 [44]

Equation:
89.50=2.35c+5.50d

If theres 22 dogs and cats in total you can plug in number less than 22 for c(# of cats) and d(# of dogs).

So we try pats numbers 8 cats and 14 dogs so

89.50=2.35(8)+5.50(14)

89.50=18.8+77

89.50=95.8

So pats numbers were wrong so we try different numbers

So i started pluging in numbers from under 22 so i got c is 10 and d is 12 lets try it with the equation

89.50=2.35(10)+5.50(12)

89.50=23.5+66

89.50=89.50 so it works the answer to how many number of dogs and cats is 12 dogs and 10 cats

6 0

3 years ago