All categories

3 years ago

Find the perimeter of a circle with radius 35cm Take(Π=22/7)

Mathematics

2 answers:

olasank [31]3 years ago

7 0

Perimeter = 2*pi*r
= 2*22/7*35
=220cm

I HOPE IT HELPED

Maslowich3 years ago

4 0

Given radius of circle = 35 cm

$\color{plum}(\pi \: value = \frac{22}{7} )$

Then, circumference (perimeter) of circle :

$= 2\pi r$

$= 2 \times \frac{22}{7} \times 35$

$= \frac{44}{7} \times 35$

$= \frac{1540}{7}$

$\color{plum}\bold{ = 220 \: cm}$

Therefore, the circumference(perimeter) of this circle = <u>220 cm</u>

You might be interested in

Cell phone company offers the followings plan: Monthly charge (which includes 400 minutes) for $30 and

xxMikexx [17]

Answer:

who are you not going to be at work at the

4 0

3 years ago

Please help meee thanks

Helga [31]

Short answer <2 and <3

Explanation
<4 is an interior angle. It has a supplementary angle in <1 which is the easiest to see. It also has another angle that is equal to it in number <5. <1 = <5 because they are vertically opposite angles. But the one that easier to see is <1.

That being said, the remote interior angles are the two interior angles that are not directly connected to <1. That is the remote part. Interior angles are those that are inside the triangle. That can only be <2 and <3.

The answer is <2 and <3 <<<< answer

5 0

3 years ago

C=r(5-d)/y We need to solve for d.

GrogVix [38]

Answer:

5-Cy/r = d

Step-by-step explanation:

C=r(5-d)/y

Multiply each side by y

Cy=r(5-d)/y *y

Cy=r(5-d)

Divide each side by r

Cy/r=r(5-d)/r

Cy/r=(5-d)

Subtract 5 from each side

Cy/r - 5 = 5-d-5

Cy/r - 5 = -d

Multiply by -1

-Cy/r + 5 = d

5-Cy/r = d

5 0

3 years ago

Read 2 more answers

how can something that could be a need become a want? A) if it is highly desired B) if the need is already fulfilled C) if it is

sergij07 [2.7K]

The answer is B) if the need is already fulfilled

8 0

4 years ago

Consider the following two ordered bases of R3:

grigory [225]

Answer:

Let $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors $b_1, ..., b_n$ at base A.

The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

at base B.

1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]$

now we use backward substitution

$c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\2\\1\end{array}\right]$

2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]$

now we use backward substitution $c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\3\\2\end{array}\right]$

3. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$