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mylen [45]
2 years ago
15

Multiply: (3√6 - √2)(4√2 - √6)​

Mathematics
1 answer:
mixer [17]2 years ago
3 0

Given multiplication:

(3√6 - √2)(4√2 - √6)

= 3√6(4√2-√6)-√2(4√2-√6)

= 3√6*4√2 - 3√6*√6 - √2*4√2 + √2*√6

= 12√12 -3√(6*6) -4√(2*2) + √(2*6)

= 12√(4*3) - 3√(36) - 4√(4) + √(12)

= 12*2√3 - 3*6 - 4*2 + 2√3

= 24√3 - 18-8 +2√3

= 24√3 + 2√3 - 18-8

= 26√3 - 18-8

= 26√3 - 26

= 26(√3-1).

<u>Answer</u><u>:</u><u>-</u> (3√6 - √2)(4√2 - √6) = 26(√3-1)

<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> Multiply. −3 2/3⋅(−2 1/4) ? −8 1/4 −6 1/6 6 1/6 8 1/4....

brainly.com/question/18369667?referrer

what is the simplified form of 6^square root of x multiplied by 6^square root of x multiplied by 6^square root of x multiplied by 6^square root of x? 1. x^2/3 2...

brainly.com/question/25224153?referrer

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(c) The total number of one centimeter lines in the first n diagrams is given by the expression
trapecia [35]

Answer:

I) f + g = 10/3

II) 4f + 2g = 20/3

III) f = 2 and g = 4/3

Step-by-step explanation:

From the chart,

P = 25

q = 40

The total number of one centimeter lines in the first n diagrams is given by the expression 

2/3n^3 + fn^2 + gn. 

When n = 1, the total number of line = 4. So,

2/3(1)^3 + f(1)^2 + g(1) = 4

2/3 + f + g = 4

Make f+g the subject of formula

f + g = 4 - 2/3

f + g = (12 - 2)/3

f + g = 10/3 ......(1)

When n = 2

Total number of line = 12

2/3(2)^3 + f(2)^2 + g(2) = 12

2/3×8 + 4f + 2g = 12

16/3 + 4f + 2g = 12

4f + 2g = 12 - 16/3

4f + 2g = (36 - 16)/3

4f + 2g = 20/3 ......(2)

(iii) To find the values of f and g, solve equation 1 and 2 simultaneously

f + g = 10/3 × 2

4f + 2g = 32/3

2f + 2g = 20/3

4f + 2g = 32/3

- 2f = - 12/3

f = 12/6

f = 2

Substitutes f in equation 1

f + g = 10/3

2 + g = 10/3

g = 10/3 - 2

g = (10 - 6)/3

g = 4/3

4 0
3 years ago
having trouble with this problem, pls help :/ I know the answer, just not how to get there. (ap calc ab, integrals)
pochemuha

Answer: D) 101

Step-by-step explanation:

By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out

\int\limits^{10}_{-1} {(2x+0.5f'(x))} \, dx =\int\limits^{10}_{-1} {2x} \, dx +0.5\int\limits^{10}_{-1} {f'(x)} \, dx =x^2|^{^{10}}_{_{-1}}+0.5f(x)|^{^{10}}_{_{-1}}\\=(100-1)+0.5(f(10)-f(-1))=99+0.5(8-4))=101

I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?

7 0
2 years ago
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Lelechka [254]
53 is the answer


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icang [17]

Answer:

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Step-by-step explanation:

x =21

y=24

<u>TH</u>=24          <u>PQ</u>=12

<u>RH</u>=24          <u>NQ</u>=12

<u>RT</u>=21            <u>NP</u>=10.5

>T=80            >P=80

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>R=80           >N=80

8 0
3 years ago
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Which of the following is an example of a proper fraction?<br> A.7/6<br> B.3/4<br> C.4/4<br> D.4/3
statuscvo [17]

Answer: B

Step-by-step explanation:

A proper fraction is when a fraction has the numeration less than the denominator

3 0
3 years ago
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