Are we supposed to find the area? If so, It's 36 because 6 times 6 equals 36 :p Hope I helped.
Answer:
Step-by-step explanation:
Sine theorem: in any triangle, the ratio between a side and the sine of the opposite angle is constant
In our case, for the left triangle,
Time to grab a calculator and crunch numbers. Double check your calculator is in degrees and not in radians (plug in sin 30°, if you're getting 0.5 you're good) and you will get
Same difference with the right triangle. With the same calculations
NOT MY WORDS TAKEN FROM A SOURCE!
(x^2) <64 => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2 so (x^2) - (8^2) < 0 To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y) You can work backwards and verify this is true. so let's set (x^2) - (8^2) equal to zero to find the roots: (x^2) - (8^2) = 0 => (x-8)*(x+8) = 0 if x-8 = 0 => x=8 and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0 so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0 so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0 so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.
Hope this helped!
:)
Do u think you could rewrite this? doesn’t really make sense
For every zero you move one decimal place to the left. there are 3 zeroes in 1000.
153.06
3 decimal places to the left
0.15306
