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ehidna [41]
3 years ago
10

Which set of side lengths could be used to form a triangle?

Mathematics
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

When the two shortest sides added up are larger then the longest side it can form a triangle

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there are 3 second grade classes at the sunshine school one has 22students and another has 20 students if there are 63 students
alex41 [277]
 there would be 21 in the last class
8 0
3 years ago
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Below are two regular polygons. Calculate the value of angle a.​
kodGreya [7K]
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7 0
2 years ago
Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
Which expression can be used to find the price of a $400 telescope after a 32% markup? Select all that apply.
elena-s [515]
<h3>There are 2 answers: choice C, choice D</h3>

====================================================

Explanation:

32% = 32/100 = 0.32

32% of 400 = 0.32*400 = 400(0.32), which adds onto the original 400 to get 400+400(0.32). This is why choice D is one of the answers.

We can factor out the GCF 400 to go from 400+400(0.32) to 400(1+0.32) which then simplifies to 400(1.32) or just 400*1.32. This shows choice C is the other answer. Using a calculator,

400+400(0.32) = 400 + 128 = 528

400*1.32 = 528

meaning that 400+400(0.32) = 400*1.32

The other answer choices result in other values, showing that they aren't equivalent to 528.

6 0
3 years ago
An artificial lake is in the shape of a rectangle and has an area of 9/20 square mile the width of the lake is 1/5 the length of
DIA [1.3K]
Answer:  The dimensions are:   " 1.5 mi.  ×  ³⁄₁₀  mi. " .
_______________________________________________
             { length = 1.5 mi. ;  width =  ³⁄₁₀  mi. } .
________________________________________________
Explanation:
___________________________________________
Area of a rectangle:

A = L * w ; 

in which:  A = Area = (9/20) mi.² ,
                L = Length = ?
                w = width = (1/5)*L = (L/5) = ?
________________________________________
  A = L * w ;  we want to find the dimensions; that is, the values for
                         "Length (L)"  and "width (w)" ; 
_______________________________________
Plug in our given values:
_______________________________________
 (9/20) mi.² = L * (L/5) ;  in which: "w = L/5" ; 
 
     → (9/20) = (L/1) * (L/5) = (L*L)/(1*5) = L² / 5 ;
   
          ↔  L² / 5  = 9/20 ;
 
            →  (L² * ? / 5 * ?) = 9/20 ?    

                →     20÷5 = 4 ;  so; L² *4 = 9 ;
 
                   ↔    4 L² = 9 ; 
 
                   →  Divide EACH side of the equation by "4" ;
           
                   →   (4 L²) / 4 = 9/4 ;
______________________________________
           to get:  →  L² = 9/4 ; 
 Take the POSITIVE square root of each side of the equation; to isolate "L" on one side of the equation; and to solve for "L" ;
___________________________________________          
 
     →   ⁺√(L²)   =   ⁺√(9/4) ;

    →   L  =  (√9) / (√4) ; 

    →  L = 3/2 ; 

    → w = L/5 = (3/2) ÷ 5 = 3/2 ÷ (5/1) = (3/2) * (1/5) = (3*1)/(2*5) = 3/10;
________________________________________________________
Let us check our answers:
_______________________________________
(3/2 mi.) * (3/10 mi.) =? (9/20) mi.² ??

→ (3/2)mi. * (3/10)mi.  =  (3*3)/(2*10) mi.² = 9/20 mi.² ! Yes!
______________________________________________________
So the dimensions are: 

Length = (3/2) mi. ;  write as: 1.5 mi.

width = ³⁄₁₀ mi.
___________________________________________________
or; write as:  " 1.5 mi.  ×  ³⁄₁₀ mi. " .
___________________________________________________
7 0
3 years ago
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