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Readme [11.4K]
3 years ago
7

Can anyone please explain this question to me?(2√5+3√6)²​

Mathematics
2 answers:
iogann1982 [59]3 years ago
7 0

\displaystyle \huge \boxed{\mathfrak{Question} \downarrow}

\displaystyle\large \bf( 2 \sqrt { 5 } + 3 \sqrt { 6 } ) ^ { 2 }

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf(  2 \sqrt { 5 } + 3 \sqrt { 6 } ) ^ { 2 }

  • Use the algebraic identify \sf\left(a+b\right)^{2}=a^{2}+2ab+b^{2}to expand \sf\left(2\sqrt{5}+3\sqrt{6}\right)^{2}.

\large \sf \: 4\left(\sqrt{5}\right)^{2}+12\sqrt{5}\sqrt{6}+9\left(\sqrt{6}\right)^{2}

  • The square of \sf\sqrt{5} is 5.

\large \sf4\times 5+12\sqrt{5}\sqrt{6}+9\left(\sqrt{6}\right)^{2}

  • Multiply 4 and 5 to get 20.

\large \sf20+12\sqrt{5}\sqrt{6}+9\left(\sqrt{6}\right)^{2}

  • To multiply \sf\sqrt{5} and \sf\sqrt{6}, multiply the numbers under the square root.

\large \sf20+12\sqrt{30}+9\left(\sqrt{6}\right)^{2}

  • The square of \sf\sqrt{6} is 6.

\large \sf20+12\sqrt{30}+9\times 6

  • Multiply 9 and 6 to get 54.

\large \sf20+12\sqrt{30}+54

  • Add 20 and 54 to get 74.

\large  \boxed{\bf12 \sqrt{30} + 74 = 139.72..}

andrey2020 [161]3 years ago
7 0

Step-by-step explanation:

<h3>(2√5+3√6)^2</h3>

<h3>(x+y) ^2= x^2+2xy+y^2, </h3>

<h3> (x = 2√5) , (y =3√6)</h3>

<h3>(2√5)^2 + 2(2√5)(3√6) + (3√6)^2</h3>

<h3>4(5) + 12√30 + 9(6)</h3>

<h3>20 + 12√30 + 54</h3>

<h3>= 74 + 12√30 or 74 + 65.72 = 139.72 </h3>

<h3>Hope, It helps You</h3>
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D.

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The formula is Sum(n) = (n -2) 180.

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Is y= 5/2x - 3 and 5x = 2y - 4 parallel or perpendicular or neither?
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Neither

Step-by-step explanation:

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If f(1) = 10 and f(n) = f(n = 1) – 4 then find the value of f(5).
svetoff [14.1K]

Given:

f(1)=10 and f(n)=f(n-1)-4.

To find:

The value of f(5).

Solution:

We have,

f(n)=f(n-1)-4

For n=2,

f(2)=f(2-1)-4

f(2)=f(1)-4

f(2)=10-4

f(2)=6

For n=3,

f(3)=f(3-1)-4

f(3)=f(2)-4

f(3)=6-4

f(3)=2

For n=4,

f(4)=f(4-1)-4

f(4)=f(3)-4

f(4)=2-4

f(4)=-2

For n=5,

f(5)=f(5-1)-4

f(5)=f(4)-4

f(5)=-2-4

f(5)=-6

Therefore, the value of f(5) is -6.

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3 years ago
Jaylon makes twice as much an hour as Sam. Sam makes x dollars an hour. Write an expression describing how much Jaylon makes an
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Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
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Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

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