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4 years ago

What is the square root of 3476

1 answer:

6 0

Square root of 346 is 58.9576118919
I hope this helps. Square root means when you multiply that number by itself it equals a number.
12^2=144
Square root of 144=12.

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Write the equation of the line parallel to the given line and passing through the given point. y= 1/2 x+2 through (0,4)

marshall27 [118]

Parallel lines have the same slope, so the slope of the new line is 1/2.

The line passes through (0,4), and this gives the y-intercept, which would equal 4.

The equation now would be: y = 1/2x + 4

3 0

3 years ago

Find the derivative of the function. y = sin−1(6x + 1)

liraira [26]

If you don't know the derivative of the inverse of sine, you can use implicit differentiation. Apply sine to both sides:

$y=\sin^{-1}(6x+1)\implies\sin y=6x+1$

(true for y between -π/2 and π/2)

Now take the derivative of both sides and solve for it:

$\cos y\dfrac{\mathrm dy}{\mathrm dx}=6$

$\dfrac{\mathrm dy}{\mathrm dx}=6\sec y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\cos\left(\sin^{-1}(6x+1)\right)}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\sqrt{1-(6x+1)^2}}$

7 0

3 years ago

I need help please.

Lera25 [3.4K]

Answer:

135

Step-by-step explanation:

first of all, the units is meters, because that is what they are asking for and have written down in the diagram.

second, to find the perimeter you must add up the values of all the sides.

as shown by the diagram all of the sides are 27 each. and there is 5 sides.

you can add 27 to itself 5 times, or you could just do 27*5. =135.

7 0

3 years ago

Q # 20 please solve the circumference

marusya05 [52]

78π in. is the answer

3 0

4 years ago

Read 2 more answers

Find the third fourth and fifth terms of the sequence defined by a1=1, a2=3, and a n =(-1)n a n-2 for n>3

Maslowich

We have been given that

$a_1=1\\ a_2=3\\
a_ n =(-1)^n a _{n-2} \text{ for } n>3$

In order to find the third term, we substitute n=3

$a_ 3 =(-1)^3a _{(3-2)}\\
\\
a_ 3 =(-1)^3 a _{1}$

Substituting the given value of $a_1$

$a_ 3 =(-1)^3 \times 1\\
\\
a_3=-1\times 1\\
\\
a_3=-1$

Similarly, we can find the fourth term by plugging n=4

$a_ 4 =(-1)^4a _{(4-2)}\\
\\
a_ 4 =(-1)^4 a _{2}\\
\\
a_4= 1\times 3\\
\\
a_4=3$

Therefore, the third term is -1

And the fourth term is 3

5 0

3 years ago