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2 years ago

Can somebody help me please

Mathematics

1 answer:

Marrrta [24]2 years ago

6 0

Answer:

(n)+(n+1)+(n+2)

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Andy played with Buzz 65% of the time. How many times did andy play with Buzz if her played with his toys 20 times?

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3 years ago

Read 2 more answers

Can someone please help me?! 100% sure please

AleksAgata [21]

3 + 3/4x > = 15
3/4x > = 15 - 3
3/4x > = 12
x > = 12 / (3/4)
x > = 12 * 4/3
x > = 48/3 = 16 sessions <==
================
-p - 4p > -10
-5p > -10
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p < 2 <===
===============
-3 - 6(4x + 6) > = 9
-3 - 24x - 36 > = 9
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=================
2x - 2 > = 10
2x > = 10 + 2
2x > = 12
x > = 12/2
x > = 6........so ur solution set is {6,7}

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2 years ago

What is 1,000 to the power of 4?

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The answer to what is 1,000 to the power of 4 is 1,000,000

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3 years ago

What is a in the quadratic eqation if the points are (3,-30), (5,0), (7,70), (9,180)

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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,

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

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<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

2 years ago

Can somebody help me please ​

Can somebody help me please