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3 years ago

Whats the square root of 81

Mathematics

2 answers:

lara31 [8.8K]3 years ago

8 0

The answer is 9 because 9x9=81. Hope this helps!

Mandarinka [93]3 years ago

3 0

Answer: It is 81.

Step-by-step explanation: If you list the multiples of 81, 9 should be in the middle. 9 X 9=81

Hope This Helps! :)

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4 years ago

A square has a diagonal that is 26 feet long.<br> What is the perimeter of this square?

Bond [772]

answer:

52 $\sqrt{2}$

step-by-step explanation:

we can use the 45-45-90 triangle theorem for this: x, x, x $\sqrt{2}$

this is because this is a square

squares have two sides of the same length, and the diagonal is different
the x's stand for the sides, whereas the x $\sqrt{2}$ stands for the diagonal

x $\sqrt{2}$ = 26

x = 13 $\sqrt{2}$

now multiply this by 4 because this is the length of one of the sides of the square. in order to find the perimeter we have to either add this four times or multiply by 4

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3 years ago

Help for this will give 5 stars and brainliest if correct and a thanks.

Soloha48 [4]

Answer:

7/4

Step-by-step explanation:

Convert the fractions into improper fractions. (7/6)*(3/2)

Then multiply across. 21/12

Simplify the fraction. 7/4

4 0

3 years ago

Read 2 more answers

Points a and b both lie on the line y=3x+7. Work out the missing y coordinate for a and the missing x coordinate for B?

Ghella [55]

Answer:

y coordinate = 13

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Step-by-step explanation:

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8 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago