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aniked [119]
3 years ago
8

E is the midpoint of DF, DE = 2x + 4, and EF = 3x + 1. Find DE, EF, and DF.

Mathematics
1 answer:
Agata [3.3K]3 years ago
8 0
When there is a midpoint in a line segment, the two segments that are then produced have to be equal. So that means that because E is a midpoint of DF, DE=EF.
Set up the equation.
DE=EF
2x+4=3x+1
Then solve for x.
x=3
Then plug it back in the equation to find DE and EF.
DE=2x+4=2(3)+4=10
Because DE=EF then EF is also 10.
DF is made from the addition of DE and EF so DF is twice EF.
DF= 20

DE= 10, EF= 10, DF= 20

Hope this helps.
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ad-work [718]

Answer:

The honey bee would go 38.85 feet if it was traveling at 10.5 feet per second for 3.7 seconds

Step-by-step explanation:

Take the speed the bee is flying (10.5) and times it by the time (3.7) and you get 38.85 which is the answer. 38.85 feet

6 0
3 years ago
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Ipatiy [6.2K]

Answer:

{2, 4, 6}

Step-by-step explanation:

the domain is the interval or set of valid x (input) values.

the range is the interval or set of valid y (result) values.

all we need to do is use the given x values in the functional definition and collect the result values. that is the range.

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7 0
2 years ago
The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
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olga nikolaevna [1]

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snow_tiger [21]

Answer:

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