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777dan777 [17]
3 years ago
15

HELP NEEDED PLEASE!

Mathematics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

the answer is about 62

Step-by-step explanation:

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ASAP HELP!! sorry the photo is bad it’s the lighting but please help
ICE Princess25 [194]

Answer:

3

Step-by-step explanation:

8 0
3 years ago
In a science test devi scores 15 marks more than lixin . If devi obtains twice as many marks as lexin . Find the number of marks
BabaBlast [244]

Answer:

Lixin obtains 15 marks

Step-by-step explanation:

Lets write two equations based on the question where: <u>D= Devi and L=Lixin</u>

D=L+15 <em>- "devi scores 15 marks more than lixin"</em>

D=2L <em>- "devi obtains twice as many marks as lexin"</em>

Now, substitute 2L as D in the first equation we wrote: (D=L+15)

2L=L+15

Next, subtract L from both sides:

<u>L=15</u>

3 0
4 years ago
What is the first step in solving the equation 2/3x+1/3+2=5?
julsineya [31]

The first step is combining like terms. On the left side of the equation add all the numbers together that don't have and x (1/3 +2)

Hope this helped! Let me know if there is anything else I can do!

7 0
3 years ago
Just fill out the explanation on why it's 6?
g100num [7]

Answer: I subtracted 8 from both sides then divided 3 on both sides

Step-by-step explanation: welcome

8 0
3 years ago
Read 2 more answers
Estimate the square root to the nearest integer and tenth square root of 8
Sever21 [200]

Answer:

√

8

≈

3

Explanation:

Note that:

2

2

=

4

<

8

<

9

=

3

2

Hence the (positive) square root of  

8

is somewhere between  

2

and  

3

. Since  

8

is much closer to  

9

=

3

2

than  

4

=

2

2

, we can deduce that the closest integer to the square root is  

3

.

We can use this proximity of the square root of  

8

to  

3

to derive an efficient method for finding approximations.

Consider a quadratic with zeros  

3

+

√

8

and  

3

−

√

8

:

(

x

−

3

−

√

8

)

(

x

−

3

+

√

8

)

=

(

x

−

3

)

2

−

8

=

x

2

−

6

x

+

1

From this quadratic, we can define a sequence of integers recursively as follows:

⎧

⎪

⎨

⎪

⎩

a

0

=

0

a

1

=

1

a

n

+

2

=

6

a

n

+

1

−

a

n

The first few terms are:

0

,

1

,

6

,

35

,

204

,

1189

,

6930

,

...

The ratio between successive terms will tend very quickly towards  

3

+

√

8

.

So:

√

8

≈

6930

1189

−

3

=

3363

1189

≈

2.828427

8 0
3 years ago
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