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3 years ago

A system of linear equations can have zero, one, or an infinite number of solutions.

2 answers:

8 0

If the linear equations form parallel lines or if the answer is not true like 0=8 or 5=18 it means the linear equations have no solutions. If the linear equations form 2 lines that are on top of each other or the answer is true for example 0=0 or 5=5 then the linear equations will have infinitely many solutions. If a linear equation gives you a single answer like x= 7 an y=9 then the linear equation only has one solution

Pani-rosa [81]3 years ago

8 0

A system of linear equations is a set of two or more linear equations in the same variables.

y=x+1 Equation 1
y=2x-7 Equation 2

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A regular hexagon and a rectangle have the same perimeter, P. A side of the hexagon is 4 less than the length, l, of the rectang

disa [49]

The answer and explanation is given in the picture below

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3 years ago

I NEED HELP ASAP PLEASE!!!!

Firdavs [7]

Answer:

Step-by-step explanation:

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2 years ago

What is the median for the following scores for Mrs. Johnson's first test? 59, 22, 87, 44, 96, 38, 70, 85, 79

katrin2010 [14]

To work out the median, you need to numerically sort the data (because currently, it's unordered).

The ordered set of data is: 22, 38, 44, 59, 70, 79, 85, 87, 96

Now you just need to pick the middle value, which is 70:

(as you can see: 22, 38, 44, 59, 70, 79, 85, 87, 96)

Therefore, your answer is 70 :)

8 0

3 years ago

Read 2 more answers

2. What unit of measure would you use for the volume of water in a bathtub?

JulsSmile [24]

2. You would use liters for a bathtub.

3.You would use milliliters for a medicine capsule.

7 0

3 years ago

Use the algebraic procedure explained in section 8.9 in your book to find the derivative of f(x)=1/x. Use h for the small number

Triss [41]

Answer:

By definition, the derivative of f(x) is

$lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Let's use the definition for $f(x)=\frac{1}{x}$

$lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}$

Then, $f'(x)=\frac{-1}{x^2}$

7 0

3 years ago