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Nataliya [291]
3 years ago
5

Determine the missing side length:

Mathematics
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

35 units

Step-by-step explanation:

Using Pythagoras Thereoem:

a² + b² = c²

21² + 28² = c²

c = √21² + 28² = √441 + 784 = √1,225 = 35 units

You might be interested in
275x=500-125x<br><img src="https://tex.z-dn.net/?f=275x%20%3D%20500%20-%20125x%20" id="TexFormula1" title="275x = 500 - 125x " a
Fynjy0 [20]
The value of x is 1.25 or 1 1/4

Explanation
==============================
275x=500-125x 
Move variable to the left side and change its sign
275x+125x=500
Collect the like terms
275x+125x=500 =====> 400x=500
Divide both sides of the equation by 400
400x/400=500 =======> x=5/4 which is the same as 1 1/4 or 1.25
Hope this helps : )
5 0
3 years ago
Trying to help my sister with her homework can some help with #2. Please and thank you so much!
pogonyaev

0.93m= 93 cm

15mm= 1.5 cm

7.5 cm

93x1.5x7.5=1046.25 cm3

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
Hi . how u r doing good. pls help me with these questions , I really need help
sleet_krkn [62]

Answer:

Which grade's question is this?

5 0
3 years ago
Draw graphs of the following lines: The line through the origin having a slope of 1/4
KengaRu [80]
I think its 1 on the graph correct me if im wrong
4 0
3 years ago
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