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3 years ago

Can someone help me plz

Mathematics

1 answer:

lukranit [14]3 years ago

8 0

The answer is area= 113.05 Circumference= 37.68

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445.76×9.634 is approximately equal to

Shalnov [3]

Answer:

Step-by-step explanation:

It sounds like you are asked to estimate it. That's a very valuable skill.

9.634 is about 10

445.76 is about 446

So an approximate answer is 446 * 10 = 4460

How close is that to the actual answer? Let's find out.4294.45

On a test, if you get an answer like 4460 in your head, and your calculator says 4294.45, you are pretty certain that you have done the question correctly.

8 0

3 years ago

The volume of a cylinder is 1280pi cm cubed. The radius is 8cm. Find the missing length.

Mademuasel [1]

Look take a picture of your problem and Brainly will get the answer for u just like that within 3 seconds

8 0

3 years ago

Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

8 0

2 years ago

Margo found that 3x741 = 2,223. How could you use mental math to find the product of 6 x 741?

kvasek [131]

Since 3 x 741 = 2223, we know 2x3= 6 so if we multiply 2 to 3 on the left side of the equation to get 6 then we can multiply a 2 to 2223 we can multiply 2 x 2223=4446.

3 0

3 years ago

Set up the appropriate trigonometric ratio to determine the value of the safety angle.

raketka [301]

Step-by-step explanation:

sin/cos=tan theta.It is the value of the safety angle

5 0

3 years ago