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Diano4ka-milaya [45]
4 years ago
8

Determine the value of x. O2 square root 6 O 6square root 2 O 6square root 3 O 12

Mathematics
1 answer:
german4 years ago
5 0

Answer:

6√2

Step-by-step explanation:

use pythagorean theorem

x=√(6^2+6^2)

or, x=√(36+36)

therefore,

x=6√2

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18=3(3x-6) multi step equation
Tems11 [23]

Answer:

x=4

Step-by-step explanation:

18=3(3x-6)

Divide each side by 3

18/3=3/3(3x-6)

6 = 3x-6

Add 6 to each side

6+6 = 3x-6+6

12 = 3x

Divide by 3

12/3 = 3x/3

4 =x

4 0
4 years ago
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Which statements are true about the graph shown below. There are more than one answer.
Galina-37 [17]

Answer:

cant see the choices what the graph is a function

Step-by-step explanation:

7 0
4 years ago
If 2(x + 2) + 4(x - 1) = 8, then 3x =
lubasha [3.4K]

Answer:

3x=4

Step-by-step explanation:

3 0
3 years ago
Part I
klemol [59]
It depends on the shape it may be rotation
6 0
4 years ago
If p is the smallest of four integers, what is their sum interms of P​
xxMikexx [17]

Answer:

If  p  is the smallest of  n  consecutive integers of the same sign than we have  p ,  p+1 ,  p+2 ,  … ,  p+(n−1) ,

So the sum is

∑k=0n−1(p+k)=∑k=0n−1p+∑k=0n−1k=np+n2−n2  

Here  n=4  

So we have  4p+6  

And checking

p+(p+1)+(p+2)+(p+3)=4p+6  

Note if  p=−v  

Than you have the same thing as if  p=v−n+1  just negative for example  3  consecutive integers the smallest is  −5  so the sum is  −5+(−4)+(−3)=3×−5+32−32=−15+3=−12  

On the other hand:

−(3+4+5)=−(3×3+32−32)=−(9+3)=−12  

If  p=−v  the sum of next  v+1  integers is  −(∑k=0vk)=−(v2+v2)  

Than needs an other  v  integers to bring it up to  0  again. From there it is

∑k=0hk=h2+h2  

Where  h=n−(2v+1) .

So recap if  p  is the smallest of  n  consecutive integers their sum is

p+(p+1)+(p+2)+…+(p+(n−1))=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪np+n2+n2−(n|p|+n2+n2)((n−|p|)2+(n−|p|)2)−(|p|p+|p|2+|p|2)((n−|p|)2+(n−|p|)2)n≥0p<0∧n<|p|+1p<0∧|p|<n<2|p|+1p<0∧n>2|p|+1

Step-by-step explanation:

6 0
3 years ago
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