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3 years ago

Calculate the value (1+√2)^4 - (1-√2)^4 without using table

Mathematics

1 answer:

nlexa [21]3 years ago

3 0

$(1 + \sqrt{2^{4} }) = 5$

$(1 - \sqrt{2^{4} } ) = - 3$ 5--3=8

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What did I do wrong?! What’s the right answer?!

skad [1K]

Answer:

The mistake was on your second step where you took b to be +6, yet it has to be -6.

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4 years ago

Find dy/dx by differentiating implicitly <br> x^2y+3xy^3-x=3

AleksandrR [38]

$\dfrac{\mathrm d}{\mathrm dx}\left[x^2y+3xy^3-x\right]=\dfrac{\mathrm d}{\mathrm dx}[3]$
$2xy+x^2\dfrac{\mathrm dy}{\mathrm dx}+3y^3+9xy^2\dfrac{\mathrm dy}{\mathrm dx}-1=0$
$(x^2+9xy^2)\dfrac{\mathrm dy}{\mathrm dx}=1-2xy-3y^3$
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3 years ago

(1−√(5))×(2−√(5)) <br><img src="https://tex.z-dn.net/?f=%281%20%2B%204%20%5Csqrt%7B10%7D%20%29%282%20-%20%20%5Csqrt%7B10%7D%20%2

Leviafan [203]

To simplify these expressions, you can either use a calculator (which I'm assuming your teacher doesn't want you to do) or FOIL (distribute).

The first one would be:

$(1- \sqrt{5}) (2- \sqrt{5}) \newline =(1)(2)+(1)(- \sqrt{5} )+(- \sqrt{5})(2)+(- \sqrt{5})(- \sqrt{5}) \newline =2- \sqrt{5}-2 \sqrt{5}+5 \newline =7-3 \sqrt{5}$

The second one would be:

$(1+4 \sqrt{10})( 2- \sqrt{10}) \newline =(1)(2)+(1)(- \sqrt{10})+(4 \sqrt{10} )(2) +(4 \sqrt{10})(- \sqrt{10}) \newline =2-\sqrt{10}+8 \sqrt{10} -40 \newline =-38+7 \sqrt{10}$

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3 years ago

There are only blue ,yellow cubes and green cubes in a bag there are 5 times as many blue cubes as yellow cubes and there are 5

Molodets [167]

Answer:

1 / 31

Step-by-step explanation:

Given that:

Yellow (y) ; Blue (b) ; Green (g)

b = 5y

g = 5b

Assume :

Number of yellow cube = 1

y = 1

b = 5

g = 5*5 = 25

Probability = (required outcome / number of total possible outcomes)

P( picking a yellow) = number of yellow / total number of cubes

Total number of balls = (1 + 5 + 25) = 31

P(picking a yellow) = 1 / 31

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3 years ago

AB BG, AC GC, ZABC LGBC ZBACZBGC

Naddika [18.5K]

Answer:

Step-by-step explanation:

1) Reflection over $\overline{BC}$

2) The triangles are congruent by SAS, so when $\triangle GBC$ is mapped onto $\triangle ABC$ , because points $G$ and $A$ are corresponding, $G'$ will correspond with $A$ .

3) Yes, because $\overline{BG} \cong \overline{AB}, \overline{GC} \cong \overline{AC}, \angle BAC \cong \angle BGC$ , so the triangles are congruent by SAS.

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2 years ago

Calculate the value (1+√2)^4 - (1-√2)^4 without using table​

Calculate the value (1+√2)^4 - (1-√2)^4 without using table