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2 years ago

12

Johnny measured the wading pool at the Belleville Community Center and calculated that it has a circumference of 6.28 meters. Wh

at is the pool's diameter?

1 answer:

Degger [83]2 years ago

6 0

Answer:

your mom

Step-by-step explanation:

im on top she is on the bottom <3

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How many ounces are in a pound

Anna35 [415]

16 ounces

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4 years ago

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A laptop carry bag cost $30. The total cost will include an 8% sales tax. Which expression can be used in calculating the total

Tamiku [17]

Answer:

30+8%=C

Step-by-step explanation:

place the cost of the bag first then add the precent and make it equal to C wich is cost

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3 years ago

Suppose there are 106 (1 million) asteroids that are 0.87 km in radius. If they were all combined into a single sphere, calculat

Eduardwww [97]

Answer:

87km

Step-by-step explanation:

Volume is proportional to radius cubed, so radius is proportional to the cube root of volume.

Increase volume by a factor of 10^6 and radius increases by a factor of the cube root of 10^6, which is 10^(6/3) = 10^2 = 100.

100 * 0.87 km = 87 km

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3 years ago

6x-3 <-9 can someone help me asap

Zanzabum

Answer:

x<-1

Step-by-step explanation:

6x-3 <-9

Add 3 to each side

6x-3+3 <-9+3

6x < -6

Divide by 6

6x/6 < -6/6

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3 years ago

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The math question is on the image<br>find the nth term of the sequence

Rus_ich [418]

Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.

Pattern 3 is Pattern 2 with 5 more balls.

Pattern 4 is Pattern 3 with 6 more balls.

Generalizing the trend, we expect Pattern $n$ to be identical to Pattern $n-1$ with $n+2$ more balls.

If $b_n$ is the number of balls in the $n$ -th pattern, then we have the recursive relation

$\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}$

We can solve this recurrence by substitution. Using the definition of $b_n$ , we have

$b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)$

$b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)$

and so on, down to

$b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)$

Recall that

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then we find

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

$\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)$

$\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2$

$\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}$

8 0

2 years ago