Answer:
C. the largest value of that class
Step-by-step explanation:
Class limit is like an interval it has the lowest value and largest value and the all values between these lower and upper values lie within that class.
Hence, the Upper Class limit of a class is the largest value of that class.
Thus, only option C is correct.
Answer:
multiply equn q by 3 .
when u will multiply equn 1 by 3 then you will get,9x+3y=39 and second equn is4x_3y=13 so, y can be cancelled and get the value of x for the first time and by putting the value of x you will find the value of y.
Answer:
5/8
Step-by-step explanation:
there are 8 slices.
if he and his friends eat 5 pieces, they ate 5 out of 8 pieces.
in maths, we write 5 out of 8 like 5/8
hope this helped, any questions - just ask
good luck,
-cheesetoasty
Given : g(x) = 
and h(x)=2x-8.
Let us find g*h(x) function now.
g*h(x) = g(x) * h(x) = 
Or g*h(x) =(2x-8).
He have square root(x-4) in composite function f*h(x).
So, we need to find the domain, we need to check for that values of x's, square root(x-4) would be defined.
Square roots are undefined for negative values.
Therefore, we can setup an inequality for it's domain x-4≥0.
Adding 4 on both sides, we get
x-4+4≥0+4.
x≥4.
Therefore, Domain is all values greater than or equal to 4.
But, restrictions would be all values less than 4 (because less than 4 would give a negative number inside square root).
Answer:
129.9 cm^2
Step-by-step explanation:
You can use the Law of Sines to find the hypotenuse. Because it's an equilateral triangle, it's the same value for the base. If you plug in the proper angles, being 90 degrees and 60 degrees because it's an equilateral triangle, the value of the base should be 17.32050808, or 17.32 if we're rounding. If you use the usual formula for finding the area of a triangle, b * h / 2 to get an answer of 129.9 cm^2. Hope this helps and isn't too confusing! There's a bunch of lessons on Khan Academy if you need more help!
This is a bit advanced for middle school, so we might be missing some information.
