All categories

3 years ago

The pizza shown has a radius of 11 cm. What is the approximate area of the pizza? (Use 3.14 for .)

Mathematics

2 answers:

Artist 52 [7]3 years ago

6 0

The approximate area of the pizza is 13cm

irga5000 [103]3 years ago

4 0

A= $\pi r^{2}$
A= 3.14* $11^{2}$
A=3.14*121
A=379.94 $cm^{2}$

You might be interested in

6. Aşağıdaki cümlelerin hangisinde “de/da”

babunello [35]

Answer:

b that is my answer : ) i hope this is correct answer

7 0

3 years ago

I don't understand this

igomit [66]

Looking at the picture attached, you will notice that on one side, we have a side of length 7+5=12. Next, we have a side length of 4+5+2=11. Hence, as length*width=area for a rectangle, this is the overall area of the rectangle. Next, in the picture, there are clearly two sides taken out - a 2x5 as well as a 4x5 rectangle, which are subtracted from the total area, resulting in 11*12-2*5-4*5
Feel free to ask further questions!

6 0

3 years ago

Eiko is wearing a magic ring that increases the power of her healing spell by 30\%30%30, percent. Without the ring, her healing

Veronika [31]

Answer:

(13/10)H

got it on khan academy brainliest pls thx uwu

Remember that 30% in fraction form is 33/100

The amount of health points (H) restored would depend on the amount of the current H so it means it would add 30% of the current which we can write as:

(30/100)H

And since it would add that to the current total we can right the current total as:

(100/100)H

So our equation would be

=(30/100)H + (100/100)H

=(130/100)H

=(13/10)H

5 0

3 years ago

Read 2 more answers

Simplify the following using PEMDAS:<br> 20 / (7-2)*3

Schach [20]

Answer:

if the * is equal to multiply, the answer is 12

Step-by-step explanation:

6 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago