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Doss [256]
3 years ago
8

Plz help. jus wanna answer :)

Mathematics
1 answer:
skad [1K]3 years ago
8 0

Answer:

<em><u>The </u></em><em><u>measure</u></em><em><u> </u></em><em><u>of </u></em><em><u>uT </u></em><em><u>is </u></em><em><u>given </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>1</u></em><em><u>.</u></em><em><u>6</u></em>

<em><u>and </u></em><em><u>it </u></em><em><u>is </u></em><em><u>given </u></em><em><u>,</u></em><em><u>that </u></em><em><u>WV </u></em><em><u>bisect</u></em><em><u> </u></em><em><u>the </u></em><em><u>line </u></em><em><u>UT</u></em>

<em><u>so,</u></em><em><u>the </u></em><em><u>measurement </u></em><em><u>of </u></em><em><u>US </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>1</u></em><em><u>.</u></em><em><u>6</u></em><em><u>/</u></em><em><u>2</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>5</u></em><em><u>.</u></em><em><u>8</u></em><em><u> </u></em>

<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> and</u></em><em><u> your</u></em><em><u> day</u></em><em><u> will</u></em><em><u> be</u></em><em><u> full</u></em><em><u> of</u></em><em><u> happiness</u></em>

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What is the slope of the line passing through (-3, 5) and (5, -3)?
Liula [17]
The slope of the line is 1.
4 0
4 years ago
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|2-5x|+3&lt;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
HELPPPPPPPPPP ASAPPPPPP!
IrinaVladis [17]
I think it’s B because of the logical math explanation
7 0
3 years ago
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. Each parent has two of these for a particular gene. an- in arty-an Off
Oliga [24]
Each parent has two ALLELES for a particular gene.
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3 years ago
C is the midpoint of segment AD. B is the midpoint of segment AC. BC = 12. What is the length of segment AD? *
guapka [62]

Answer:

48

Step-by-step explanation:

A - B - C - - - D

 12 + 12 + 24 = 48

Segment B is midpoint between AC therefore the length between AB is equal to the length between BC.  C is midpoint of of AD therefore the length between AB + BC is equal to the length between CD.  IF AB is 12 then AB is 12.  AC will be 24.  Since CD is equal to AD, CD must also be 24.  AD is distance AC plus CD = 48.

7 0
3 years ago
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