Answer:
x = -6
Step-by-step explanation:
Solve for x:
-x/3 = 2
Multiply both sides of -x/3 = 2 by -3:
(-3 (-x))/3 = -3×2
-3×(-1)/3 = (-3 (-1))/3:
(-3 (-1))/3 x = -3×2
(-3)/3 = (3 (-1))/3 = -1:
--1 x = -3×2
(-1)^2 = 1:
x = -3×2
-3×2 = -6:
Answer: x = -6
Answer:Use proportions to find scale measurements or actual ... If a drawing is to scale, you can use proportions to determine the ... First, create a proportion
Step-by-step explanation:
Use proportions to find scale measurements or actual ... If a drawing is to scale, you can use proportions to determine the ... First, create a proportion
Answer:
y = 2 + ((-)6/5)x
Step-by-step explanation:
-6x-5y=-10
add 6x to both sides.
-5y = -10 +6x
divide both sides by -5
y = 2 - (6/5)x
Plug in 0 for x to get the y intercept:
f(0) = 2 - (6/5) (0)
y = 2
(0, 2) is the y intercept.
Do the same for values such as -1, -2, 1, and 2, etc.
Then graph it.
Since only half of the 8 white eggs are cracked, then there are 4 cracked, white eggs. The probability of drawing one of these 4 from the 18 total eggs is 4/18 which simplifies to 2/9 or 22%.
I hope this was quick enough haha, have a nice day and remember that Jesus loves you <3
Answer:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Step-by-step explanation: