Answer:
1. So, it's basically y=mx + b, you put the b first like y= 2/3x -1, and the b is negative, so you go down on the y-axis one time and the 2/3x is rise over run 2 is the rise and 3 is the run.
for y=-x -4 you put a denominator for -x like -1/1x you keep the denominator positive.
2. If the 2x is in the incorrect spot you put it on the other side and it's a positive, so you put is as a negative on the other side, so you're left with
-y=-2x-1, but you can't have a sign on them, so you divide it with -1 and on the other side as well, so it comes down to y=2x+1 and if the x doesn't have a denominator you add a one to it, like this y=2/1x+1 and you put the 1 on the y-axis and its a positive so you go up one time and the rise or run is 2/1, so you start at 1 on the y-axis and you rise 2 times, so 3-axis and run 1 time and it's 1-axis.
For the last one ill do it on paper.
Answer:
for number one the answer is no for 7 the answer is 5 seconds
Step-by-step explanation:
Answer: n= -20?
Step-by-step:
Simplify both sides of the equation
1/5n+4=0
Then subtract 4 from both sides
1/5n + 4 - 4 = 0 - 4
1/5n= -4
Then multiply both sides by 5
5*(1/5n)=5*(-4)
Answer:
No.
Step-by-step explanation:
It has different order of matrices .
For <em>A</em><em>d</em><em>d</em><em>i</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em>or <em>S</em><em>u</em><em>b</em><em>s</em><em>t</em><em>r</em><em>a</em><em>c</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em>, both matrices must have the same number of <u>r</u><u>o</u><u>w</u><u>s</u> and <u>c</u><u>o</u><u>l</u><u>u</u><u>m</u><u>n</u><u>s</u> .