All categories

1 year ago

Id love some help!!!

Mathematics

2 answers:

timofeeve [1]1 year ago

3 0

Answer : 12.56 inches²
Explanation :
Diameter is twice the radius.
Radius = 4 ÷ 2 = 2 inches
Area = 3.14r² = 3.14 × 2² = 12.56 inches²

Mariulka [41]1 year ago

3 0

Answer:

<h2>Circle:</h2>

circle is a special kind of ellipse in which the eccentricity is zero and the two foci are coincident.

$\sf{formula \: to \: find \: the \: area \: is}$

$\large{ \boxed{ \tt{A = {\pi \times r}^{2} }}}$

$\sf{formula \: to \: find \: perimeter \: of \: circle \: is}$

$\large {\boxed{ \tt{p = 2 \times \pi \times r}}}$

Continue to Question.

If the radius is 2 inches and

Use 3,14 for π So:

$\tt{ {A = \pi \times {r}^{2} }}$

$\tt{A = (3.14 \times 2 \times 2)}$

$\red{ \boxed{ \bold{A = 12.56 {inches}^{2} }}}$

So, The Area of circle is 12.56inches²

You might be interested in

I don't understand a single thing on how to get an answer. you know how to answer it?

Ainat [17]

Answer:

it is 6^3

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Write the given linear system in matrix form. Assume X = x y z . dx dt = −8x + 7y − 9z dy dt = 5x − y dz dt = 10x + 7y + 8z

Akimi4 [234]

Answer:

The matrix form of given linear system is

$\begin{bmatrix}x'\\ y'\\ z'\end{bmatrix}=\begin{bmatrix}-8 & 7 &-9\\ 5 & -1 & 0\\ 10 & 7 & 8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}$

Step-by-step explanation:

Let as assume that

$X=\begin{bmatrix}x\\ y\\ z\end{bmatrix}$

Given differential equations are

$\frac{dx}{dt}=-8x+7y-9z$

$\frac{dy}{dt}=5x-y$

$\frac{dz}{dt}=10x+7y+8z$

We need to find the matrix form of given linear system.

Write the elements of left side in a column matrix.

Write all the coefficients in one matrix first which is called a coefficient matrix. Multiply coefficient matrix with the variables matrix and equate left and right side.

$\begin{bmatrix}\frac{dx}{dt}\\ \frac{dy}{dt}\\ \frac{dz}{dt}\end{bmatrix}=\begin{bmatrix}-8 & 7 &-9\\ 5 & -1 & 0\\ 10 & 7 & 8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}$