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Colt1911 [192]
3 years ago
15

Line segment EF has a length of 5 units. It is translated 5 units to the right on a coordinate plane to obtain line segment E′F′

. What is the length of E′F′?
5 units
6 units
10 units
25 units
Mathematics
2 answers:
mylen [45]3 years ago
7 0

Answer:

25 units

Step-by-step explanation:

just multiply by 5

Dimas [21]3 years ago
3 0

Answer 25 units

Step-by-step explanation:

cause there are already 5 so multipy 5

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Derivative of this function<br> <br>Y= -7X³ -8X² +6X +3
zaharov [31]

In general, the derivative of a single term  Ax^(n)  is  A n x^(n-1) .

And the derivative of a sum of many terms is the sum of the derivatives
of the individual terms.  

Using these two rules, the derivative (with respect to 'x') of the expression
in the question is . . .

<em>            Y' = -21x² - 16x + 6</em>


7 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
IF U HELP RN ILL GIVE U BRAINLIEST
shutvik [7]

Answer:

i need more info. that part dont help. if you took pic of the sentence they gv you that would be easier to work with.

Step-by-step explanation:

5 0
3 years ago
Please help me with this problem​
AleksandrR [38]

Answer:

The answer to your question is Area = 6 u²

Step-by-step explanation:

Process

To find the area of the picture divide the figure between a square and a right triangle, determine the area of each figure and finally add them up.

1.- Calculate the area of the square

Area = side x side

        = 2 x 2

        = 4 u²

2.- Calculate the area of the triangle

Area = base x height / 2

base = 2   height = 2

Area = 2 x 2 /2

Area = 2 u₂

3.- Total area

Area = 4 + 2

       = 6 u²

3 0
4 years ago
Could someone pls tell me the answer Thankss
Anni [7]
I think 10,000 not sure though
7 0
4 years ago
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