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Shalnov [3]
3 years ago
13

Which situation could be described by this expression?

Mathematics
2 answers:
Dvinal [7]3 years ago
4 0
25 divided by a number will produce a smaller number and equal parts of the numerator. which answer would result in smaller equal parts?

B
ioda3 years ago
4 0
Your answer will be b hope that helps
You might be interested in
A cola container is in the shape of a right circular cylinder. The radius of the base is 4 centimeters, and the height is 10 cen
noname [10]

Answer:

V=160\pi

Step-by-step explanation:

V=h\pi r^{2}

V=10*\pi 4^{2}

V=160\pi

5 0
3 years ago
Lified expression for 2(X-5)?
Marina CMI [18]

Answer:

This, simplified would be 2x-10

8 0
3 years ago
How do you solve them
Andrew [12]
#1)     9=6 - 3 |4p + 9| now we're going to distribute
          
          9= 6 - 12p - 27
         
          9= 6 - 27 - 12p
          
          9= - 21 - 12p
  
          9 + 21= -12p
   
          30= - 12p  Now you are going to divide both side by -12

         -12:30 = -12:-12p

          -5/2= p final answer

#2)  - |1 + 4a| = -13   now you going to change the sign
        
        -1 - 4a = -13

        -4a = -13 + 1

        -4a = -12  now divide both side by -4 

        -4a:-4 = -12:-4

a = 3 final answer









4 0
3 years ago
Sam and Bethan share £54 in the ratio 5:4. how much will each person get
Sunny_sXe [5.5K]
Given :- 
sam and Bethan shareare in the ratio 5:4 respectively. 
 Total no of money is = 54ÂŁ 
 Solution :-
 Let Sams share is be x ie = 5 
 Let Bethans share be y ie = 4
 Total no of share is = 9
 Therefore the persons share is (particular persons share / total no of share) *
total value of money 
 Ans : -
 Therefore sams share x is = (5/9)* 54 =30ÂŁ
 Therefore Bethans share y is = (4/9) * 54 = 24ÂŁ
5 0
3 years ago
Using principle of mathematical induction prove that 6^-1 divisble by 5 .​
salantis [7]

I suppose the claim is 5 \mid 6^n - 1 for n\in\Bbb N.

When n=1, we have 6^1 - 1 = 6 - 1 = 5, and of course 5 divides 5.

Assume the claim holds for n=k, that 5 \mid 6^k - 1. We want to use this to show it holds for n=k+1, that 5 \mid 6^{k+1} - 1.

We have

6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5

Since 5 \mid 6^k - 1, we can write 6^k - 1 = 5\ell for some integer \ell. Then

6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)

which is clearly divisible by 5. QED

5 0
2 years ago
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