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Oksanka [162]
2 years ago
14

HELPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!! in assigned photo is the question! doesnt need to be descriptive i just need the answer ple

ase, im so behind

Mathematics
2 answers:
aev [14]2 years ago
8 0

Answer: 11

Step-by-step explanation:

Since the smaller triangle is half of the larger triangle

5.5×2=11

Art [367]2 years ago
8 0

Answer:

The correct answer is 11.

Step-by-step explanation:

You might be interested in
Please Help me with this I've been trying to figure this out​
grandymaker [24]

Answer:

angle FXG is congruent to angle IXJ or JXI. GHI is 81°

Step-by-step explanation:

Angle FXG is congruent to angle IXJ because the two are vertical angles. (Vertical angle theorem)

FXG + GXH = FXH

Substitute the given values given:

49 ° + GXH = 130°

Subtract 49 both sides

49 - 49 + GXH = 130 - 49

GXH = 81°

7 0
3 years ago
|2-5x|+3<10 how to solve
ivann1987 [24]
1 Simplify — 2 <span>Equation at the end of step  1  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  2  :</span> 3 Simplify —— 10 <span>Equation at the end of step  2  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  3  :</span> 2 Simplify — 5 <span>Equation at the end of step  3  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  4  :</span>Calculating the Least Common Multiple :

<span> 4.1 </span>   Find the Least Common Multiple

      The left denominator is :      <span> 5 </span>

      The right denominator is :      <span> 10 </span>

<span><span>        Number of times each prime factor
        appears in the factorization of:</span><span><span><span> Prime 
 Factor </span><span> Left 
 Denominator </span><span> Right 
 Denominator </span><span> L.C.M = Max 
 {Left,Right} </span></span><span>5111</span><span>2011</span><span><span> Product of all 
 Prime Factors </span>51010</span></span></span>


      Least Common Multiple:
      10 

Calculating Multipliers :

<span> 4.2 </span>   Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 2

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

<span> 4.3 </span>     Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent, <span> y/(y+1)2  </span> and <span> (y2+y)/(y+1)3  </span>are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

<span> L. Mult. • L. Num. 2x • 2 —————————————————— = —————— L.C.M 10 R. Mult. • R. Num. 3 —————————————————— = —— L.C.M 10 </span>Adding fractions that have a common denominator :

<span> 4.4 </span>      Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

2x • 2 - (3) 4x - 3 ———————————— = —————— 10 10 <span>Equation at the end of step  4  :</span> (4x - 3) 1 ———————— - — = 0 10 2 <span> Step  5  :</span>Calculating the Least Common Multiple :

<span> 5.1 </span>   Find the Least Common Multiple

      The left denominator is :      <span> 10 </span>

      The right denominator is :      <span> 2 </span>

<span><span>        Number of times each prime factor
        appears in the factorization of:</span><span><span><span> Prime 
 Factor </span><span> Left 
 Denominator </span><span> Right 
 Denominator </span><span> L.C.M = Max 
 {Left,Right} </span></span><span>2111</span><span>5101</span><span><span> Product of all 
 Prime Factors </span>10210</span></span></span>


      Least Common Multiple:
      10 

Calculating Multipliers :

<span> 5.2 </span>   Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = 5

Making Equivalent Fractions :

<span> 5.3 </span>     Rewrite the two fractions into equivalent fractions

<span> L. Mult. • L. Num. (4x-3) —————————————————— = —————— L.C.M 10 R. Mult. • R. Num. 5 —————————————————— = —— L.C.M 10 </span>Adding fractions that have a common denominator :

<span> 5.4 </span>      Adding up the two equivalent fractions

(4x-3) - (5) 4x - 8 ———————————— = —————— 10 10 <span> Step  6  :</span>Pulling out like terms :

<span> 6.1 </span>    Pull out like factors :

   4x - 8  =   4 • (x - 2) 

<span>Equation at the end of step  6  :</span> 4 • (x - 2) ——————————— = 0 10 <span> Step  7  :</span>When a fraction equals zero :<span><span> 7.1 </span>   When a fraction equals zero ...</span>

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

4•(x-2) ——————— • 10 = 0 • 10 10

Now, on the left hand side, the <span> 10 </span> cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   4  •  (x-2)  = 0

Equations which are never true :

<span> 7.2 </span>     Solve :    4   =  0

<span>This equation has no solution.
</span> A a non-zero constant never equals zero.

Solving a Single Variable Equation :

<span> 7.3 </span>     Solve  :    x-2 = 0<span> 

 </span>Add  2  to both sides of the equation :<span> 
 </span>                     x = 2

One solution was found :                   x = 2

<span>
</span>

4 0
3 years ago
A square's side is 5.1 m. Find its area
natima [27]
5.1 times 4 sides equals the answer would be 20.4<span />
8 0
3 years ago
Use a quadratic equation to find two real numbers that satisfies the situation. The sum of the two numbers is 7 and their produc
vovangra [49]
The answer is D. No such number exists. You would get and imaginary solution.
3 0
3 years ago
Select all the conditions for which it is possible to construct a triangle. A triangle with side lengths 4 cm, 5 cm, and 15 cm A
maxonik [38]

9514 1404 393

Answer:

  • A triangle with side lengths 3 cm. 5 cm. and 7 cm
  • A triangle with side lengths 4 cm, 5 cm. and 6 cm.​

Step-by-step explanation:

The sum of the short two sides must exceed the length of the longest side. That is not the case for lengths {4, 5, 15} or {5, 5, 10}.

Triangles can be formed from {3, 5, 7} and {4, 5, 6}.

3 0
3 years ago
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