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2 years ago

Prove the sum of two rational numbers is rational where a, b, c, and d are integers and b and d cannot be zero.

Mathematics

1 answer:

timama [110]2 years ago

5 0

Answer:

hahahhaa thanks

Step-by-step explanation:

ahahahhahahad

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NEED HELP PLEASE ITS URGENT <br> thanks- BRAINLIST only for correct answers love :)<br> Thank you!!

Rashid [163]

Answer:

true flase true

Step-by-step explanation:

8 0

2 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

Plz hellppppp me u won’t regret it

nexus9112 [7]

Answer:

Area of circle = 200.96 units²

Step-by-step explanation:

The formula to find the area of a circle is: πr², where r is the radius (which is half of the diameter)

So the diameter is 16 and of 16 is 8, which is the radius.

Sub it into the formula of a circle and you would get the area of a circle.

3.14 x 8²

Use calculator

Area of circle = 200.96 units²

8 0

2 years ago

what is the answer to this it takes sara 1/2 day to make birdhouse how much of birdhouse will be built after 2/5 day

Fiesta28 [93]

Answer:80%

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Quadratic equations and complex numbers PLEASE HELPPPP ASAP

kondor19780726 [428]

we are given

$27x^3-1=0$

we can also write it as

$(3x)^3-(1)^3=0$

now, we can use factor formula

$a^3-b^3=(a-b)(a^2+ab+b^2)$

we can use above formula

we get

$(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)$

now, we can simplify it

$27x^3-1=(3x-1)(9x^2+3x+1)$

now, we can set it to 0

$27x^3-1=(3x-1)(9x^2+3x+1)=0$

and then we can solve for x

$(3x-1)(9x^2+3x+1)=0$

$3x-1=0$

$x=\frac{1}{3}$

$9x^2+3x+1=0$

now, we can use quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

now, we can plug values

and we get

$x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}$

$x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$

So, we will get solution as

$x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$ ...............Answer

5 0

2 years ago