Answer:
7.6
Step-by-step explanation:
From the graph, we know the lengths of two legs.
   a = 3
   b = 7
Knowing this, we can use the Pythagorean Theorem to solve for c, the hypotenuse. (remember,  a^2 + b^2 = c^2)
c =	√a2 + √b2  
=	√(7)2 + √(3)2
=	√49 + √9
=	√58
=	7.6
 
        
             
        
        
        
Answer:
64.01
Step-by-step explanation:
tan∅= p/b
tan58= x/40
or, x= 40×tan58
 x = 64.01
 
        
             
        
        
        
The sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5
Given,
18th term of an arithmetic sequence = 8.1
Common difference = d = 0.25.
<h3>What is an arithmetic sequence?</h3>
The sequence in which the difference between the consecutive term is constant.
The nth term is denoted by:
a_n = a + ( n - 1 ) d 
The sum of an arithmetic sequence:
S_n = n/2 [ 2a + ( n - 1 ) d ]
Find the 18th term of the sequence.
18th term = 8.1
d = 0.25
8.1 = a + ( 18 - 1 ) 0.25
8.1 = a + 17 x 0.25
8.1 = a + 4.25
a = 8.1 - 4.25
a = 3.85
Find the sum of 20 terms.
S_20 = 20 / 2 [ 2 x 3.85 + ( 20 - 1 ) 0.25 ]
          = 10 [ 7.7 + 19 x 0.25 ]
          = 10 [ 7.7 + 4.75 ]
          = 10 x 12.45
          = 124.5
Thus the sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5
Learn more about arithmetic sequence here:
brainly.com/question/25749583
#SPJ1
 
        
             
        
        
        
After using the slope formula your answer would be 5/0.1 but when simplifying your answer is 50
        
             
        
        
        
There are  ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
 ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
There are  ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's
 ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's  way of doing that.
 way of doing that.
So we have

The last expression has a more compact form in terms of the so-called multinomial coefficient,
