Let
x--------> the number

we know that
11²=121
12²=144

then
x² must be greater than 121 and less than 144

case a) √115
if x=√115
then
x²=115-------> is not greater than 121

case b) √121
if x=√121
then
x²=121-------> is not greater than 121

case c) √136
if x=√136
then
x²=136-------> is greater than 121 and is less than 144------> is ok

case d) √150
if x=√150
then
x²=150------> is not less than 144

therefore

the answer is
√136

8 0

3 years ago

Read 2 more answers

Identify the square root of the following. * √196

Fynjy0 [20]

Answer:

Step-by-step explanation:

5 0

3 years ago

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The number of vibrations n n per second of a nylon guitar string varies directly with the square root of the tension T T and inv

emmasim [6.3K]

Answer: $T=40.96\ KgF$

Step-by-step explanation:

We know that:

$n$ : The number of vibrations per second of the nylon guitar string.

$T$ : Tension.

$L$ : The length of the string.

Since $n$ varies directly with the square root of the $T$ and inversely with $L$ , the equation has the following form:

$n=k*\frac{\sqrt{T} }{L}$

Where "k" is the constant of variation.

Knowing that when $n=15$ and $L=0.6$ , $T=256$ , we can find the value of "k":

$n=k*\frac{\sqrt{T} }{L}\\\\L*n=k\sqrt{T}\\\\\frac{L*n}{\sqrt{T}}=k\\\\k=\frac{(0.6)(15)}{\sqrt{256}}\\\\k=0.5625$

Finally, in order to find the tension when the length is 0.3 meters and the number of vibrations is 12, you need to substitute these values and the value of "k" into $n=k*\frac{\sqrt{T} }{L}$ and solve for $T$ :

$12=(0.5625)\frac{\sqrt{T} }{0.3}\\\\\frac{12(0.3)}{(0.5625)}=\sqrt{T}\\\\(6.4)^2=T\\\\T=40.96\ KgF$

8 0

3 years ago