Ask question

Ask question

All categories

4 years ago

6

HELP PLEASE What is the following product?

2 answers:

Naya [18.7K]4 years ago

6 0

The answer would be:

sergeinik [125]4 years ago

3 0

$(2 \sqrt{7}+3 \sqrt{6} )(5 \sqrt{2}+4 \sqrt{3} )=10 \sqrt{14}+8 \sqrt{21} +15 \sqrt{12}+12 \sqrt{18} = \\ =10 \sqrt{14}+8 \sqrt{21} +15 *2\sqrt{3}+12*3 \sqrt{2}= \\ = 10 \sqrt{14}+8 \sqrt{21} +30\sqrt{3}+36 \sqrt{2}$

You might be interested in

Which polynomial identity will prove that 49-4=45

goldfiish [28.3K]

Consider the polynomial identity (difference of squares):
(x+y)(x-y) = x² - y²

Set x =7 and y = 2 to obtain
(7+2)*(7-2) = 7² - 2²
9*5 = 49 - 4
45 = 49 - 4
This is the result required, obtained by using difference of squares.

Answer: Difference of squares.

3 0

4 years ago

Question Navigator<br> Enter the answer.cubed roots What is

Helen [10]

The answer is 11..and there's a pic for u too...my pleasure helping you

5 0

4 years ago

558 cups of water fill 412 identical water bottles. How many cups fill each bottle ?

Elan Coil [88]

Answer:

5/4 cups fill each bottle

Step-by-step explanation:

5 5/8=45/8

4 1/2=9/2

Step by Step:

------------------

(45/8)/(9/2)

(45/8)(2/9)

(45/4)(1/9)

45/36

5/4

8 0

3 years ago

Sales tax she paid I . L . N 2-

goldenfox [79]

Rounded it equals $30. It was really like $29.83 :)

4 0

4 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago