Answer:
r≤0
Step-by-step explanation:
if the range is s≤0, (that means the value of s is all negative integers &0) to find the domain we have to replace that some negative integers &0.
we can replace some values of s as the following
s=‒4(0)^2 s=‒4(‒1)^2
=‒4(0) =‒4(1)
=0 =‒4
s=‒4(‒2)^2 s=‒4(‒3)^2
=‒4(4) =‒4(9)
=‒16 =‒36
so we can understand easly that the domain is negative integer&0,it can't be positive because the square of any integer is positive.
therefore the answer is r≤0
i think &i hope it is clear.
Answer:
x = 
Step-by-step explanation:
The segment from the vertex to the base is a perpendicular bisector.
Using Pythagoras' identity on the right triangle on the left.
x² = 2² + 3² = 4 + 9 = 13 ( take the square root of both sides )
x = 
Answer:
x=-5
Step-by-step explanation:
-2x=3+7
or,-2x=10
or,x=-10/2
x=-5
36100/36789=0.9812715757427492
Well, the answer quite simple ....there are 4! ways to arrange these numbers...and as 4! = 4*3*2*1 = 24
hence 24 is the correct answer....
u can also remember it by theorem of multiplication as.....in first place (I.e. first code can be any no. out of 4,5,2&7 ....so 4*.....
as first place is acquired by a certain no. that leaves three no. to fill third place and when third place I'd occupied it leaves 2 numbers to fill second place and lastly only one no. to fill the last place .....so it's result will be like 4*3*2*1.
I know this is pretty much confusing ....but still I tried my best....
if anything troubles u here feel free to ask me