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leva [86]
3 years ago
10

Will mark brainlest how to multipy it​

Mathematics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

\displaystyle x   = 2 \sqrt{3}

Step-by-step explanation:

we would like to solve the following equation:

\displaystyle  \sqrt{3}  =  \frac{6}{x}

in order to do so do cross multiplication:

\displaystyle x \sqrt{3}  =  6

divide both sides by √3:

\displaystyle x   =   \frac{6}{ \sqrt{3}}

since we ended up with a square root on the denominator so we can consider rationalising the denominator to do so multiply both numerator and denominator by √3 which yields:

\displaystyle x   =   \frac{6}{ \sqrt{3}}  \times  \frac{ \sqrt{3} }{ \sqrt{ 3} }

simplify multiplication:

\displaystyle x   =   \frac{6 \sqrt{3} }{ 3}

reduce fraction:

\displaystyle x   = 2 \sqrt{3}

and we are done!

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Arrange the integers 1,4,8,12,17,32 so that all consecutive integers add to a perfect square. Enter your answer with numbers sep
viktelen [127]

One possible answer is: 1, 8, 17, 32, 4, 12

I don't know if there's a methodical way to get this answer. I used trial and error. I started with 1 and looked through the list to see what adds to 1 to get a perfect square. That would be 8 since 1+8 = 9 = 3^2

The process is repeated but this time for 8. After a bit of guess and checking, we see that 8+17 = 25 = 5^2.

Then after 17 is 32 because 17+32 = 49 = 7^2.

Keep doing this until all of the values are used up. If you get stuck, then try backtracking to a previous branch/path where everything worked and try another fork in the road. If that doesn't work, then try starting the sequence with a completely different value (instead of 1).

Other answers may be possible. I haven't checked all possible permutations.

6 0
2 years ago
Five rectangles are arranged from the least to the greatest area and named A, B, C, D, and E in order of increasing area. All di
Hatshy [7]

Given that median area is 15 square units.

Hence rectangle C in the middle has 15 square units.

Its dimensions can be width= 5 and length = 3

SInce B is smaller than C and has the same length, B has lengh of 3 with area = 9 sq units.

D has the same perimeter = 16 units.  Since D is a square, side of D = 4 units.

Now D and E have the same length.  Hence length of E = 4 units.

Width of E = width of C = 5 units.Thus makes the area of E as 20 sq units.

Rectangle A has length =3 and width can be less than 3 since area is smaller than B.

so A has length= 3 width = 2 with area = 6 sq units.

8 0
2 years ago
Find the general term of {a_n}
Assoli18 [71]

From the given recurrence, it follows that

a_{n+1} = 2a_n + 1 \\\\ a_{n+1} = 2(2a_{n-1} + 1) = 2^2a_{n-1} + 1 + 2 \\\\ a_{n+1} = 2^2(2a_{n-2}+1) + 1 + 2 = 2^3a_{n-2} + 1 + 2 + 2^2 \\\\ a_{n+1} = 2^3(2a_{n-3} + 1) + 1 + 2 + 2^2 = 2^4a_{n-3} + 1 + 2 + 2^2 + 2^3

and so on down to the first term,

a_{n+1} = 2^na_1 + \displaystyle \sum_{k=0}^{n-1}2^k

(Notice how the exponent on the 2 and the subscript of <em>a</em> in the first term add up to <em>n</em> + 1.)

Denote the remaining sum by <em>S</em> ; then

S = 1 + 2 + 2^2 + \cdots + 2^{n-1}

Multiply both sides by 2 :

2S = 2 + 2^2 + 2^3 + \cdots + 2^n

Subtract 2<em>S</em> from <em>S</em> to get

S - 2S = 1 - 2^n \implies S = 2^n - 1

So, we end up with

a_{n+1} = 4\cdot2^n + S \\\\ a_{n+1} = 2^2\cdot2^n + 2^n-1 \\\\ a_{n+1} = 2^{n+2} + 2^n - 1 \\\\\implies \boxed{a_n = 2^{n+1} + 2^{n-1} - 1}

5 0
2 years ago
PLEASEEEEEE HELPPPPP!!
DaniilM [7]

Answer:

y-b+am/m

Step-by-step explanation:

y-b=m(x-a)

y-b=MX-ma

mx=y-b+ma

x=y-b+ma/m

4 0
2 years ago
Bradley bought 14 yellow highlighters. each highlighter cost $0.94.how much did Bradley spend?
Svetradugi [14.3K]
The answer would be 13.16$

0.94x14=13.16
6 0
3 years ago
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