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satela [25.4K]
3 years ago
12

Select all the equations that represent the distance formula. Group of answer choices LaTeX: d=\sqrt{\left(x_1-x_2\right)^2+\lef

t(y_1-y_2\right)^2} d = ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 LaTeX: d=\sqrt{\left(x_2+x_1\right)^2+\left(y_2+y_1\right)^2} d = ( x 2 + x 1 ) 2 + ( y 2 + y 1 ) 2 LaTeX: d=\sqrt{\left|x_2-x_1\right|^2+\left|y_2-y_1\right|^2} d = | x 2 − x 1 | 2 + | y 2 − y 1 | 2 LaTeX: d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 LaTeX: d=\sqrt{\left(x_2+x_1\right)-\left(y_2+y_1\right)}
Mathematics
1 answer:
grigory [225]3 years ago
7 0

Answer:

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Step-by-step explanation:

The distance between two points A(x₁,y₁) B(x₂,y₂) is given by the formula as follows :

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

It is used to find the distance between two points. It is an application of the Pythagorean theorem. Usually, the distance covered is equal to the product of speed and time. But when speed and time in not given but coordinates are given, then only we can use this formula.

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Solve the linear system if 3x+4y-z=-6 5x+8y+2z=2. -x+y+z=0
Anni [7]
We are given with three equations and three unknowns and we need to solve this problem. The solution is shown below:
Three equations are below:
3x + 4y - z = -6
5x + 8y + 2z = 2
-x + y + z = 0

use the first (multiply by +2) and use the second equation:
2 (3x+4y -z = -6)  => 6x + 8y -2z = -12
                               + ( 5x + 8y +2z = 2)
                                  ------------------------
                                     11x + 16y = -10  -> this the fourth equation

use the first and third equation:
 3x + 4y -z = -6
+ (-x + y + z =0)
-------------------------
        2x + 5y = -6 -> this is the fifth equaiton

use fourth (multiply by 2) and use fifth (multiply by -11) equations such as:
2 (11x + 16y = -10)   => 22x + 32y = -20  -> this is the sixth equation
-11 (2x + 5y = -6)    => -22x -55y = 46   -> this is the seventh equation

add 6th and 7th equation such as:
   22x + 32y = -20
+(-22x - 55y = 66)
---------------------------
             - 23y = 46
               <span>   y = -2

solving for x, we have:
</span>2x + 5y = -6 
2x = -6 - 5y
2x = -6 - (5*(-2))
2x = -6 +10
2x = 4
x=2

solving for y value, we have:
-x + y + z =0
z = x -y 
z = 2- (-2)
z =4

The answers are the following:
x = 2
y = -2
z = 4
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3 years ago
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Answer:

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Answer:

Company A: $530; Company B: $610; Difference: $80

Step-by-step explanation:

Company A:

12(40) = 480 + 50 = 530

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4 0
3 years ago
The function below has at least one rational zero.
Dmitry_Shevchenko [17]

Answer:

x=-7,x=-1,x=-0.289898,x=0.689898

or x=-7,x=-1,x=\frac{1}{5}-\frac{\sqrt{6}}{5},x=\frac{1}{5}+\frac{\sqrt{6}}{5}

Step-by-step explanation:

<u>Factor by grouping</u>

g(x)=5x^4+38x^3+18x^2-22x-7\\\\g(x)=5x^4+33x^3-15x^2-7x+5x^3+33x^2-15x-7\\\\g(x)=x(5x^3+33x^2-15x-7)+1(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+35x^2-2x^2-14x-x-7)\\\\g(x)=(x+1)(5x^2(x+7)-2x(x+7)-1(x+7))\\\\g(x)=(x+1)(x+7)(5x^2-2x-1)\\\\0=(x+1)(x+7)(5x^2-2x-1)\\

We can easily see that the first two zeroes are x=-1 and x=-7, however, the third zero will need the help of the quadratic formula:

0=5x^2-2x-1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-1)}}{2(5)}\\\\x=\frac{2\pm\sqrt{4+20}}{10}\\ \\x=\frac{2\pm\sqrt{24}}{10}\\ \\x=\frac{2\pm2\sqrt{6}}{10}\\ \\x=\frac{2}{10}\pm\frac{2\sqrt{6}}{10}\\ \\x=\frac{1}{5}\pm\frac{\sqrt{6}}{5}\\\\x_1\approx0.689898\\\\x_2\approx-0.289898

Therefore, the zeroes of the function are x=-7,x=-1,x=-0.289898,x=0.689898

7 0
3 years ago
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