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cupoosta [38]
4 years ago
13

When solving the equation 4(3x^2 +2)-9=8x^2+7. Emily wrote 4(3x + 2)=8x^2 + 16 as her first step. Which property justifies Emily

's first step? State a reason why1) Addition Property of Equality2)Commutative Property of Addition3) Multiplication Property of Equality4) Distributive Property of Multiplication Over Addition
Mathematics
1 answer:
vfiekz [6]4 years ago
6 0
Hello : 
4(3x² +2)-9=8x²+7 = <span>4(3x^2 +2)-9 +9 =8x²+7 +9 
                                = </span><span> 4(3x² + 2)=8x² + 16
</span><span>(State a reason why1) Addition Property of Equality)</span>
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Substitute this equation x+3y=18 &amp; -x-4y=-25
sveticcg [70]

Answer:

 {x,y} = {3/7,-43/7}

Step-by-step explanation:

System of Linear Equations entered :

  [1]    x + 3y = -18

  [2]    -x + 4y = -25

Graphic Representation of the Equations :

   3y + x = -18        4y - x = -25  

 

 

Solve by Substitution :

// Solve equation [1] for the variable  x  

 

 [1]    x = -3y - 18

// Plug this in for variable  x  in equation [2]

  [2]    -(-3y-18) + 4y = -25

  [2]    7y = -43

// Solve equation [2] for the variable  y  

  [2]    7y = - 43  

  [2]    y = - 43/7  

// By now we know this much :

   x = -3y-18

   y = -43/7

// Use the  y  value to solve for  x  

   x = -3(-43/7)-18 = 3/7  

Solution :

{x,y} = {3/7,-43/7}

5 0
3 years ago
Why do I cry when I listen to juice wrld?
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8.008
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The digit 7 is in the tens place so 74 will become either 70 or 80 after rounding off to the nearest ten. Draw a number line with 3 numbers on it: 70, 75 and 80. (Both possible rounded off numbers and the number in the middle.
3 0
3 years ago
let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep
Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

x_1,x_2 and x_3 are linearly independent.

By definition of linear independent there exits three scalar a_1,a_2 and a_3 such that

a_1x_1+a_2x_2+a_3x_3=0

Where a_1=a_2=a_3=0

y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1

We have to prove that y_1,y_2 and y_3 are linearly independent.

Let b_1,b_2 and b_3 such that

b_1y_1+b_2y_2+b_3y_3=0

b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0

b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0

(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0

b_1+b_3=0

b_1=-b_3...(1)

b_1+b_2=0

b_1=-b_2..(2)

b_2+b_3=0

b_2=-b_3..(3)

Because x_1,x_2 and x_3 are linearly independent.

From equation (1) and (3)

b_1=b_2...(4)

Adding equation (2) and (4)

2b_1==0

b_1=0

From equation (1) and (2)

b_3=0,b_2=0,b_3=0

Hence, y_1,y_2 and y_3 area linearly independent.

5 0
3 years ago
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