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iVinArrow [24]
2 years ago
5

What is the eye ball shaped as?

Mathematics
1 answer:
tia_tia [17]2 years ago
4 0

Answer:

the eyes ball are some as D . MARBLE .

there are round.

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Can someone help with this question, please?
riadik2000 [5.3K]

Answer:

the answer is D

and here a way to get the correct

and it is to

keep,change the sign,and flip the last fraction

Step-by-step explanation:

3 0
3 years ago
utilizando la equivalencia indicada, resuelve el siguiente ejercicio: 1 pulgada = 2.54cm 4.2 cm = _____ pulgadas
Phoenix [80]
<span>Los 4.2 centímetros es equivalente a 1.6 pulgadas</span>
4 0
4 years ago
If set A has 4 elements and B has 3 elements then set n(AXB) is a) 12 b) 14 c) 24 d) 7
Karo-lina-s [1.5K]

Answer:

The correct option is a.

Step-by-step explanation:

Given information:

Number of elements in set A = 4

Number of elements in set B = 3

We need to find n(A × B). It means we have to find the number of elements in set A × B.

If P and Q are two sets with elements m and n respectively, then the number of element in the set P × Q is m × n.

n(A\times B)=n(A)\times n(B)

n(A\times B)=4\times 3

n(A\times B)=12

The value of n(A × B) is 12. Therefore the correct option is a.

8 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)

so

\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}

so

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)

so

\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
4 years ago
12. Karen earns $20 for every three cake she sells. Which graph models a relationship with the same unit rate?​
Kruka [31]

Answer:

Graph B

Step-by-step explanation:

There is a clear relationship between the amount of money she earns and the cakes she sells

5 0
4 years ago
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