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marusya05 [52]
3 years ago
7

Name a pair of perpendicular lines: is line BC and AJ perpendicular: explain ​

Mathematics
1 answer:
wel3 years ago
8 0
A pair of perpendicular lines would be line CF and line AJ.

Line BC and line AJ are NOT perpendicular because when a pair of lines are perpendicular, all angles become 90 degrees. It is obvious, in the case, if you were to slide line BC on top of line AJ, the angles will not equal 90 degrees.
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Paul needs 214 2 1 4 yards of fabric to make a tablecloth. How many tablecloths can he make from 1812 18 1 2 yards of fabric?
ladessa [460]

Answer:

Paul can make 0

Step-by-step explanation:


6 0
3 years ago
Which ordered pair is a solution of the equation y = 10x?<br><br>​
ddd [48]

Answer:

THE ANSWER WILL BE...

NOTE : Just substitute the x and y variable with (-6,-60) and you will see I am right.

Example:

-60= 10(-6)

<em><u>Please mark as brainliest</u></em>

Have a great day, be safe and healthy

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6 0
3 years ago
Consider the system of equations.
Harman [31]

The solution to the system of equations x + 2y = 1  and -3x-2y = 5 is:

x  = -3,  y = 2

The given system of equations:

x  +  2y  =  1............(1)

-3x - 2y  =  5..........(2)

This can be written in matrix form as shown:

\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]

Find the determinant of \left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right]

\triangle = 1(-2) - 2(-3)\\\triangle = -2+6\\\triangle  = 4

\triangle_x = \left[\begin{array}{ccc}1&2\\5&-2\end{array}\right]\\\triangle_x = 1(-2)-2(5)\\\triangle_x = -2-10\\\triangle_x =-12

\triangle_y = \left[\begin{array}{ccc}1&1\\-3&5\end{array}\right]\\\triangle_y = 1(5)-1(-3)\\\triangle_y = 5 + 3\\\triangle_y =8

x = \frac{\triangle_x}{\triangle} \\x = \frac{-12}{4} \\x = -3

y = \frac{\triangle_y}{\triangle} \\y = \frac{8}{4} \\y = 2

The solution to the system of equations x + 2y = 1  and -3x-2y = 5 is:

x  = -3,  y = 2

Learn more here: brainly.com/question/4428059

3 0
2 years ago
Solve the equation using the Zero­Product Property. 6x(3x + 1) = 0
neonofarm [45]
   
\displaystyle  \\ &#10;6x(3x + 1) = 0 \\  \\ &#10;\Longrightarrow ~~ 6x = 0~~\texttt{or}~~3x+1=0 \\  \\ &#10;6x = 0 ~~\Longrightarrow ~~ x_1 =  \frac{0}{6} = \boxed{0} \\  \\ &#10;3x+1 = 0 ~~\Longrightarrow ~~ 3x = -1  ~~\Longrightarrow ~~ x_2 =  \frac{-1}{3} = \boxed{-\frac{1}{3}} \\  \\



4 0
3 years ago
Without solving the equation, determine the nature of the roots of x^2 - 2x - 8 = 0.
noname [10]

Answer:

two real, unequal  roots

Step-by-step explanation:

This is a quadratic equation.  The quadratic formula can be used to determine how many and what kind of roots may exist:  

Find the discriminant, which is defined as b^2 - 4ac, if ax^2 + bx + c = 0.  In this case, a = 1, b = -2 and c  = -8, so that the discriminant value is

(-2)^2 - 4(1)(-8), or 4 + 32  =  36.

Because the discriminant is real and positive, we know for certain that we have two real, unequal  roots

3 0
3 years ago
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