How to classify a system and give the number of solutions

1 answer:

3 0

1) consistent system. at least one solution.

2) inconsistent system. no solutions (the lines are parallel)

3) dependent system. equations have same slope and y-intercept(same line)independent system.

4) equations have different slopes(2 intersecting lines)

im not sure if thats what you were looking for

You might be interested in

The difference of two times a number and five is eight more than the number

Finger [1]

Add it then x it by 3

4 0

4 years ago

Given: The coordinates of isosceles trapezoid JKLM are J(-b, c), K(b, c), L(a, 0), and M(-a, 0). Prove: The diagonals of an isos

In-s [12.5K]

We have to find the lengths of the diagonals KM and JL:
d ( KM ) = √ (( - a - b )² + ( 0 - c )²) = √ (( a + b )² + c² )
d ( JL ) = √ ( ( a - ( - b ) )² + ( 0 - c )²) = √ ( ( a + b )² + c² )
So the lengths of the diagonals KM and JL are congruent.
The lengths of the diagonals of the isosceles trapezoid are congruent.

6 0

3 years ago

Read 2 more answers

((12+5)+20÷4))+4<br>order of operations

likoan [24]

((12+5)+20÷4))+4

first is parentheses inside out

12+5 = 17

((17)+20÷4))+4

now divide inside the parenthesis

(17+5)+4

next parentheses

22+4

Answer: 26

5 0

3 years ago

Read 2 more answers

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

3 years ago

I need to know the answer my sister keeps bugging me and i do not know

solong [7]

This football field has 36 squares. As each square's area is 1 m^2. So, total area is 36 m^2.
the area = 8×1 = 8 m^2( this portion is not shaded )
i.e, the area of shaded part will be (36-8) = 28 m^2.

6 0

3 years ago