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exis [7]
3 years ago
5

Use Euler’s Formula to find the missing number. Edges: 43 Faces: 25 Vertices: ?

Mathematics
2 answers:
Ymorist [56]3 years ago
7 0
The answer is D, 23.
babunello [35]3 years ago
6 0

Answer:

Vertices = 20

Step-by-step explanation:

Using the Euler's formula:

V - E + F = 2            ....[1]

where,

E is the number of Edges

V is the number of Vertices

F is the number of Faces.

As per the statement:

Edges: 43  

Faces: 25  

⇒E = 43 and F = 25

Substitute in [1] we have;

V-43+25 = 2

⇒V-18 = 2

Add 18 to both sides we have;

V = 20

Therefore, the number of vertices is, 20

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Simplify and write as a decimal.
oksian1 [2.3K]

11.9

i just got it right on a quiz so yeah

6 0
3 years ago
This 1 seems really complicated
Fofino [41]
The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
________________________________________________________
Given: 
________________________________________________________
 y = - 4x + 16 ; 

 4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ; 

→  4y − x + 4 = 0 ; 

Subtract "4" from each side of the equation ; 

→  4y − x + 4 − 4 = 0 − 4 ;

→  4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________
   
y = - 4x + 16 ;   ===> Refer to this as "Equation 1" ; 

4y − x =  -4 ;     ===> Refer to this as "Equation 2" ; 
________________________________________________________
Solve for "x" and "y" ;  using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→  " y = - 4x + 16 " ;
_______________________________________________________
→  Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

       to solve for "x" ;   as follows:
_______________________________________________________
Note:  "Equation 2" :

     →  " 4y − x =  - 4 " ; 
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y";  in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→   as follows:  
_________________________________________________

→   " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

   a(b + c)  = ab + ac ;   AND: 

   a(b − c) = ab <span>− ac .
_________________________________________________
As such:

We have:  
</span>
→   " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→    "4 (-4x + 16) "  =  (4* -4x) + (4 *16)  =  " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→  " -16x + 64 − x = - 4 " ; 

Note:  " - 16x − x =  -16x − 1x = -17x " ; 

→  " -17x + 64 = - 4 " ;   Solve for "x" ; 

Subtract "64" from EACH SIDE of the equation:

→  " -17x + 64 − 64 = - 4 − 64 " ;   

to get:  

→  " -17x = -68 " ;

Divide EACH side of the equation by "-17" ; 
   to isolate "x" on one side of the equation; & to solve for "x" ; 

→  -17x / -17 = -68/ -17 ; 

to get:  

→  x = 4  ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ; 

which is:  " y = -4x + 16" ; to solve for "y".
______________________________________

→  y = -4(4) + 16 ; 

        = -16 + 16 ; 

→ y = 0 .
_________________________________________________________
The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ; 
_________________________________________________________
→  Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten; 

→  Let us check;  

→  For EACH of these 2 (TWO) equations;  do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ; 

→ Consider the first equation given in our problem, as originally written in the system of equations:

→  " y = - 4x + 16 " ;    

→ Substitute:  "4" for "x" and "0" for "y" ;  When done, are both sides equal?

→  "0 = ?  -4(4) + 16 " ?? ;   →  "0 = ? -16 + 16 ?? " ;  →  Yes!  ;

 {Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→  " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

 {that is:  "4" for the "x-value" ; & "0" for the "y-value" ;  

→  to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→    " 4(0)  −  4 + 4 = ? 0 ?? " ;

      →  " 0  −  4  + 4 = ? 0 ?? " ;

      →  " - 4  + 4 = ? 0 ?? " ;  Yes!
_____________________________________________________
→  As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→   "x = 4" and "y = 0" ;  or; write as:  [0, 4]  ;  are correct.
_____________________________________________________
Hope this lenghty explanation is of help!  Best wishes!
_____________________________________________________
7 0
3 years ago
Pls help me it would be nice !!!
notsponge [240]

Answer:

y=−3x−10

That is the equation in slope intercept form.

5 0
3 years ago
If the area of AABC is D, give the expressions that complete the equation for the measure of ZB?
Ira Lisetskai [31]

Given:

The figure of triangle ABC.

The area of the triangle ABC is D.

m\angle B=\sin ^{-1}(\dfrac{m}{n})

To find:

The value of m and n in the given expression.

Solution:

Let h be the height of the triangle ABC.

Area of a triangle is:

Area=\dfrac{1}{2}\times base\times h

Where, b is the base and h is the height of the triangle.

Area=\dfrac{1}{2}\times a\times h

The area of the triangle ABC is D.

D=\dfrac{1}{2}\times a\times h

2D=ah

\dfrac{2D}{a}=h                  ...(i)

In a right angle triangle,

\sin \theta =\dfrac{Perpendicular}{Hypotenuse}

\sin B =\dfrac{h}{c}

\sin B =\dfrac{1}{c}\times \dfrac{2D}{a}              [Using (i)]

\sin B =\dfrac{2D}{ac}

m\angle B =\sin ^{-1}\dfrac{2D}{ac}            ...(ii)

We have,

m\angle B=\sin ^{-1}(\dfrac{m}{n})          ...(iii)

On comparing (ii) and (iii), we get

m=2D

n=ac

Therefore, the required values are m=2D, n=ac.

6 0
3 years ago
How can estimating the quotient help you check that your answer to a division problem is reasonable
kap26 [50]
Round ur quotient and the divsor like 100 divide by 9... would be about 11 and 9 is about 10 and 11 is about 10 so 10 times 10 is one hundred ao the quotient is reasonable
5 0
3 years ago
Read 2 more answers
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