Answer:
1) x³y²
2) 6x²y³
Step-by-step explanation:
hope it helps...
Answer:
(a) Slope of Line 1 = 3
Slope of Line 2 = 3
Slope of Line 3 = -1/3
(b)
Line 1 and Line 2 = Parallel
Line 1 and Line 3 = Perpendicular
Line 2 and Line 3 = Perpendicular
Step-by-step explanation:
Please mark as brainliest! Thanks! Hope this helps!
First question:
When you bisect an angle, you create two new equal angles, both measuring half of the original angle. So, if the original angle was 52°, the two angles will be 52/2=26 degrees each.
Note: the picture is not very clear, and I'm not sure the original angle is 52°. It it's not, just divide by 2 whatever measure the angle is.
Second question:
Focus on triangles RQP and RSP:
- They are both right, because Q and S are right angles
- We are given QP=SP
- They share the side RP
So, they are congruent. This implies that angle SRP is 50° as well, and then
QRS = QRP+SRP = 50°+50°=100°
Cos x = 5 / 9 ==》 Cos x = 0.55
==》 x = 56 degrees
Well, there are three different ways to solve systems of equations:
a) Graphing
b) Substitution
c) Elimination
Graphing:
If your system of equations are written in slope-intercept form (y=mx+b "m" being the slope and "b" being the y-intercept) its quite easy.
The y-intercept is pretty much where your line touches the y-axis so if your first equation is: y=3x-2 then you will put your first point on (0,-2). Then from that point you follow the slope. if the slope is 3 then you'll rise 3 and run 1 (go three times up and one to the right and put your point right there) till you feel you have enough points to make a line. You do the same process with your second equation.
Substitution:
This is best when you have a system of equations in which an equation tells you what x or y are equal to. For example: if I have "4x+2=y" and "2y+3x= 26" you can "replace" your "2y" with the equation "4x+2=y" which gives you a brand new equation: " 2(4x+2)+3x=26". To solve this you do the distributive property and link like terms which gives you the equation "11x+4=26". Then you solve like a normal equation, isolating x and then you'll get the value of x (in this case x=2). Now you go to the first equation "4x+2=y" and replace "x" to its value (in this case it's 2) and you get y (in this example, y=10).
Elimination:
I think this one is one of the easiest methods. Let's say my systems of equations is "5x-6y=-32" & "3x+6y=48". If you take a look at y's coefficients, you see they are the same. So you can "cancel them out" like this:
5x-6y=-32
+ 3x+6y=48
-----------------------
8x=16 <-------- when you get this, you can solve like a normal equation
x=2
Also, if you have, for example:
2x+6y=190 & 2x+3y=130
You can solve this by: canceling x or multiplying the second equation by 2 and canceling y.
I'm gonna show you how to do the second way (multiplying and canceling y)
My second equation is "2x+3y=130". To get an equivalent equation, I just have to multiply the whole equation by any number. If I want to cancel y, then I have to multiply by 2.
So "2x+3y=130" turns into "4x+6y=260"
Now you follow the same procedure as before:
2x+6y=190
- 4x+6y=260
-------------------------------
-2x= -70
2x=70--------> x=35
Then you pretty much follow the substitution method by replacing "x" to "35" in one of the equations.
Hope this covers all you missed in class. If you have any questions you can comment them here or message me. :D Good luck in your exam!!!
