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4 years ago

Hi can you run answers or help with answers

Mathematics

1 answer:

schepotkina [342]4 years ago

3 0

1). The inequality says "height is 54 or more".
Any height that is less than 54 does NOT qualify.

2). The sign says "Maximum total weight 800 pounds".
Any weight that's 800 pounds or less is OK.

3). Paula has scored 21, 25, and 15 points.
. . . . . they're NOT all greater than 21
. . . . . they ARE all 25 or less
. . . . . they ARE all less than 30
. . . . . they ARE all 10 or more .

4). First, write down the number of cans that
each student collected, but arrange them in order.
Then check each choice.
A). misses the highest 3 students completely
B). includes everybody
C). misses the lowest 3 students completely
D). OK, but the first and last intervals aren't needed,
because nobody falls in those

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The answer is 340.687 liters

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3 years ago

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

snow_tiger [21]

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

Need f(2):

f(2)=f(2-1)+4
f(2)=f(1)+4
f(2)=(3)+4=7

FIND f(3):
f(3)=f(3-1)+4
f(3)=f(2)+4
f(3)=(7)+4
f(3)=11

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3 years ago

Does anyone know an equation for this table:

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3 years ago

What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

 $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

 $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

 $a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

 $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
 $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

4 0

3 years ago

Make vthe subjject of the formulaE=mvsquare/ 2 find the value of v when m =2 and E=64

Maksim231197 [3]

E=mv²

divide both sides by m

E/m=v²

square root both sides

v=√E/m

m=2,E=64

v=√64/2

v=9/√2

6 0

3 years ago