25.3 is what i got for the first image.
Answer:
the desired equation is y = (-1/3)x + 6
Step-by-step explanation:
Let the given line be A: y=3x+2
The slope of line A is m = 3.
The slope of any line B which is perpendicular to line A is the negative reciprocal of the slope of A: m = -(1/3).
The particular perpendicular line that passes through (3, 5) is then
5 = (-1/3)(3) + b, which simplifies to 5 = -1 + b, or b = 6.
Thus, the desired equation is y = (-1/3)x + 6
Do the second one hope this helps
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!
:)
Answer:
17
Step-by-step explanation:
So, this is a percentage problem.
Start off by finding how many students 0.28% is:
If 100% = 5780
0.01% = 0.578
Now:
0.01% = 0.578
0.28% = 16.184
The exercise tells you to round for a whole person, so 16.184 turns 17
And that's the answer!
