All categories

3 years ago

What is the side length,s, of the square?? A= 81

Mathematics

1 answer:

spin [16.1K]3 years ago

4 0

Answer:

40.5

Step-by-step explanation:

A square has 4 sides that means that you have to divide the Area by 4.

81/4= 40.5

All sides of the square will have the side lengths of 40.5.

You might be interested in

Compare 9.36 and 9.359

Sati [7]

Try this option (this is not the only way!):

1. the rule: if 'number_1' - 'number_2' > 0, then number_1>number_2. If 'number_1'-'number_2'<0, then number_2>number_1.

2. according to the rule above: 9.36-9.359=0.001. 0.001>0, it means, 9.36>9.359.

6 0

3 years ago

3(×+2)=24 can you help me

trapecia [35]

Answer:

x = 6 if that's what your asking

Step-by-step explanation:

please give brainliest

3 0

3 years ago

Read 2 more answers

Savannah has a granola Recipe that requires 1/3 pound of oats to make one batch of granola.If savannah uses 2/5 Pounds of oats w

VMariaS [17]

Answer:

only 1 batch

Step-by-step explanation:

4 0

3 years ago

HRW CAN U TRY I GIVE BRAINLYS.....

AlekseyPX

Answer:
i think the answer is C

8 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

4 years ago

What is the side length,s, of the square?? A= 81​

What is the side length,s, of the square?? A= 81