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3 years ago

What is the apparent solution to the system of equations?

Mathematics

1 answer:

Leto [7]3 years ago

8 0

Step-by-step explanation:

The given equations are :

y=-x+2 ....(1)

y = -4x +7 ....(2)

From equation (1) and (2)

-x+2 = -4x +7

Taking like terms together

-x+4x=7-2

3x = 5

x = 1.667

Put the value of x in equation (1)

y=-(1.667)+2

y = 0.333

The solution of the equations are (1.667,0.333). The attached figure shows the graph for the equations.

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Step-by-step explanation:

The number in front of x is always the slope in a y=Mx+b equation and the other number is the y intercept

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4 years ago

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Find the distance from point A to the given line.

VMariaS [17]

Answer:

Let's see what to do buddy...

Step-by-step explanation:

_________________________________

Step (1)

To find the distance between the line which formed like :

$ax + by + c = 0$

and the point which is like :

$h = \binom{m}{n} \\$

we use following equation :

$d = \frac{ |a(m) + b(n) + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } \\$

_________________________________ Step (2)

$y = - x + 4$

$x + y - 4 = 0$

So ;

$a = 1$

$b = 1$

$c = - 4$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

A = ( 2 , - 1 )

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

So we have :

$d = \frac{ | \: 1(2) + 1( - 1) - 4| }{ \sqrt{ {1}^{2} + {1}^{2} } } = \\$

$d = \frac{ |2 - 1 - 4| }{ \sqrt{1 + 1} } = \frac{ | - 3| }{ \sqrt{2} } \\$

$d = \frac{3}{ \sqrt{2} } \\$

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

5 0

4 years ago

Please help! Correct answer only, please! Consider the matrix shown below: Find the determinant of the matrix Q. A. -67 B. -65 C

Nonamiya [84]

Answer: d) 67

<u>Step-by-step explanation:</u>

$determinant\ \left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&j\end{array}\right] = a\cdot det\left[\begin{array}{cc}e&f\\h&j\end{array}\right] -\ b\cdot det\left[\begin{array}{cc}d&f\\g&j\end{array}\right] +\ c\cdot det\left[\begin{array}{cc}d&e\\g&h\end{array}\right]$

$determinant\ \left[\begin{array}{ccc}2&3&4\\-3&2&1\\5&-1&6\end{array}\right] \\\\\\= 2\cdot det\left[\begin{array}{cc}2&1\\-1&6\end{array}\right] -\ 3\cdot det\left[\begin{array}{cc}-3&1\\5&6\end{array}\right] +\ 4\cdot det\left[\begin{array}{cc}-3&2\\5&-1\end{array}\right]\\\\\\=2[2(6)-1(-1)]-3[-3(6)-1(5)]+4[3(-1)-2(5)]\\\\\\=2(13)-3(-23)+4(-7)\\\\\\=26+69-28\\\\\\=\large\boxed{67}$

8 0

4 years ago