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notka56 [123]
3 years ago
7

Why is the square root of 36 equal to 6?​

Mathematics
2 answers:
slavikrds [6]3 years ago
7 0

Answer:

6^2= 36

Step-by-step explanation:

Vadim26 [7]3 years ago
5 0

Answer:

because that is what sqrt are

Step-by-step explanation:

A square root of a number is a value that, when multiplied by itself, gives the number. Example: 6 × 6 = 36, so a square root of 36 is 6 ... The symbol is √ which always means the positive square root.

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26=4v+3v+12 can anyone help?
motikmotik

Answer:

Step-by-step explanation:

Combine like terms

3v + 4v

26=7v+12

Subtract    12  from both sides of the equation

26-12

12-12

14=7v

Simplify

so subtract the 12's

Divide both sides of the equation by the same term

14 divided by 7

7 divided by 7

Simplify

do the division and move the term on the other side and it would be

v=2

Solution

 

v=2

4 0
3 years ago
Read 2 more answers
1400000000000000000000km in scientific notation
viva [34]

1 400 000 000 000 000 000 000 km = 1.4 * 10^21 km

8 0
3 years ago
Match each operation involving f(x) and g(x) to its answer
eimsori [14]

Answer:

(g-f) (-1)= sqrt(15)

(f/g)(-1)= 0

(g+f)(2)=sqrt(3)-3

(g*f)(2)=-3*sqrt(3)

Step-by-step explanation:

We have to eval the expressions given in the point indicated.

Lets start by the first equation

(g-f)(-1)= g(-1) - f(-1)= \sqrt{11-4*(-1)}    - 1 +(-1)^{2} = \sqrt{15}

Now, lest continue with the others

(f/g)(-1)= f(-1)/g(-1)= (1-1)/sqrt(15)=0

(g+f)(2)=g(2)+f(2)=sqrt(3)-3

(g*f)(2)=g(2)*f(2)=sqrt(3)*(-3)=-3sqrt(3)

3 0
3 years ago
Read 2 more answers
Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract
anyanavicka [17]

Answer:

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8

Step-by-step explanation:

The sum you are trying to understand is this.

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}

Remember that in general when you have a geometric series  

\sum\limits_{n = 0}^{\infty} a*r^n you have that

\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}      and that equality is true as long as     |r| < 1.

Therefore here we have

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n        and   \big|\frac{-1}{15} \big| = \frac{1}{15} < 1

Therefore we can use the formula and

\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n =  \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16  = 15/8

5 0
3 years ago
Which of the following points is not collinear with (3, 6) and (-2, -9)?
Lilit [14]
Its B, cause it’s collinear
5 0
3 years ago
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