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2 years ago

Use 3.14 for pi. ROUND YOUR FINAL ANSWER TO THE NEAREST TENTHS PLACE!

Mathematics

2 answers:

Natasha2012 [34]2 years ago

7 0

The area of the given circle is 95 square ft.

Calculation for the Area of the Circle:

It is given in the diagram that,

The diameter of the circle, d = 11 ft.

Now, we know that the radius of a circle is equal to half of the diameter.

Therefore, radius of the given circle, r = 11/2 ft.

The formula for the area of the circle is given as follows,

A = π × r²

Substituting the values, π = 3.1, and r = 11/2, we get,

A = (3.14) × (11/2)²

A = (3.14) × 30.25

A = 94.985

A ≈ 95

Hence, the area of the given circle with diameter 11 ft. comes out to be 95 ft².

Learn more about a circle here:

brainly.com/question/11833983

#SPJ1

ivolga24 [154]2 years ago

4 0

Answer:

379.94

Step-by-step explanation:

3.14 x 11^2 = 379.94

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Can someone plz help me. There is an equation and the cost too.

bonufazy [111]

Answer:

C = 40p

Cost for 15 people: 600 dollars

Step-by-step explanation:

The amusement park charges $40 per person.

That can be represented as 40 × P or 40P.

C is equal to the cost of admission for P people.

So, C = 40P

C = 40P

C = 40(15)

C = 600

The cost of admission for 15 people is 600 dollars.

Hope that helps.

5 0

3 years ago

Find the value of the determinant using the method of expansion by minors; expand on the third row

valkas [14]

For the matrix

$\begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}$

the determinant using the method of expansion by minors, expanding on the third row is:

$\det \begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}=a_{31}\det \begin{bmatrix}{a_{12}} & {a_{13}} & {} \\ {a_{22}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}-a_{32}\det \begin{bmatrix}{a_{11}} & {a_{13}} & {} \\ {a_{21}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}+a_{33}\det \begin{bmatrix}{a_{12}} & {a_{12}} & {} \\ {a_{22}} & {a_{22}} & {} \\ {} & {} & {}\end{bmatrix}$

Answer:

First, we compute the determinants of the minors:

$\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}$

Therefore: