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Art [367]
3 years ago
14

Which relation is a function? A) x y = y2 - 3 B) y x = y2 - 3 C) x2 + y2 = 5 + x D) x2 - y = 10x + 25 + y

Mathematics
1 answer:
marin [14]3 years ago
7 0
Since you didn’t specify that it needed to be y as a function of x, all of the above equalities can be rewritten as functions. However, in case that you meant to imply that, A and B are both not valid for y as a simple function of x.
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What is the value of the expression shown below? Plssssss helppppp
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I think it’s 8 but unsure
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2 years ago
Without solving it, determine the nature of the roots of this equation:
ycow [4]
<h2>Answer:</h2>

x = 1, -4

Step-by-step explanation:

Find the roots of  x:

2  +  3

x  −  4  =  0  by solving for  x

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6 0
3 years ago
Anyoneee?? Mean Median and mode
juin [17]

Answer:

mode is the most occuring number which is in this case 11

you get the median by putting them in order from least to greatest and them marking them out until you get to the middle but if there are two numbers in the middle you add them and then divide by two and that is your median but in this case your median is 11

you get the mean by adding all the numbers together and dividing by how many numbers there are so in this case you are dividing by 9 and when you add oall of these numbers together you get 98 and 98/9 is 10.8888888889 which you would round and get 10.9

hope this helps

3 0
3 years ago
Rationalise the denominator of:<br>1/(√3 + √5 - √2)​
Paul [167]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} }

can be re-arranged as

\rm :\longmapsto\:\dfrac{1}{ \sqrt{3}   -   \sqrt{2}   +  \sqrt{5} }

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }  \times \dfrac{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }

We know,

\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) =  {x}^{2} -  {y}^{2} \: }}

So, using this, we get

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{ {( \sqrt{3}  -  \sqrt{2} )}^{2}  -  {( \sqrt{5}) }^{2} }

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{3 + 2 - 2 \sqrt{6}   - 5}

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{5 - 2 \sqrt{6}   - 5}

\rm \:  =  \: \dfrac{ \sqrt{3} -  \sqrt{2}   -  \sqrt{5} }{ - 2 \sqrt{6}}

\rm \:  =  \: \dfrac{ - ( -  \sqrt{3} +  \sqrt{2}  + \sqrt{5}) }{ - 2 \sqrt{6}}

\rm \:  =  \: \dfrac{-  \sqrt{3} +  \sqrt{2}  + \sqrt{5}}{2 \sqrt{6}}

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{-  \sqrt{3} +  \sqrt{2}  + \sqrt{5}}{2 \sqrt{6}}  \times \dfrac{ \sqrt{6} }{ \sqrt{6} }

\rm \:  =  \: \dfrac{-  \sqrt{18} +  \sqrt{12}  + \sqrt{30}}{2  \times 6}

\rm \:  =  \: \dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}

\rm \:  =  \: \dfrac{-  3\sqrt{2} + 2 \sqrt{3}   + \sqrt{30}}{12}

Hence,

\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} } =\dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h3><u>More Identities to </u><u>know:</u></h3>

\purple{\boxed{\tt{  {(x  -  y)}^{2} =  {x}^{2} - 2xy +  {y}^{2}}}}

\purple{\boxed{\tt{  {(x   +   y)}^{2} =  {x}^{2} + 2xy +  {y}^{2}}}}

\purple{\boxed{\tt{  {(x   +   y)}^{3} =  {x}^{3} + 3xy(x + y) +  {y}^{3}}}}

\purple{\boxed{\tt{  {(x - y)}^{3} =  {x}^{3} - 3xy(x  -  y) -  {y}^{3}}}}

\pink{\boxed{\tt{  {(x + y)}^{2} +  {(x - y)}^{2} = 2( {x}^{2} +  {y}^{2})}}}

\pink{\boxed{\tt{  {(x + y)}^{2}  -  {(x - y)}^{2} = 4xy}}}

6 0
3 years ago
Round 59.09 to the nearest whole number​
MariettaO [177]

Answer:

59

Step-by-step explanation:

It goes to 59 because the number in the decimals place is a 0, for 0-4 you round down and for 5-9 you round up.

8 0
3 years ago
Read 2 more answers
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