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3 years ago

Tangle JKL similar to triangle PQR. Solve fort T

Mathematics

1 answer:

lubasha [3.4K]3 years ago

5 0

We need a picture to see what to solve

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3 years ago

A candy company has 150 kg of chocolate-covered nuts and 90 kg of chocolate-covered raisins to be sold as two different mixes. O

Westkost [7]

Answer:

1. 7x + 9.5y

2. 1980

Step-by-step explanation:

In Mix A, there is an equal amount of nuts and raisins.

Let x = kg of nuts = kg of raisins.

In Mix B, there is 3 times as much nuts as raisins.

Let y = kg of raisins, then 3y = kg of nuts.

We have this information:

Nuts Raisins Price

mix A x + x = $ 7.00

mix B 3y + y = $ 9.50

Total: 150 90

Hence, the number of kilograms of each mix should company prepare for the maximum revenue is = 7x + 9.5y

We have these inequalties:

{x≥0y≥0x+3y≤150x+y≤90} (equation 1)

There will be

x+x=2x kg of Mix A at $7.00 per kg.

The revenue from Mix A is:

7(2x)=14x dollars.

There will be

3y+y=4y kg of Mix B at $9.50 per kg.

The revnue from Mix B is:

9.5(4y)=38y dollars.

Hence, the total revenue is:

R=14x+38y (equation 2).

Graph the region determined by [1] and locate its vertices.

You should get:

(0,0),(90,0),(0,50),(60,30)

Test the vertices into [2] and see which vertex produces maximum revenue

(0,0): R=14(0)+38(0)=0 Minimum revenue

(90,0):R=14(90)+38(0)=1260

(0,50): R=14(0)+38(50)=1900

(60,30):R=14(60)+38(30)=1980 Maximum revenue.

3 0

4 years ago

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