Suppose h, a, h,a, and bb are positive real numbers, and t_0, y_0t 0 ,y 0 are real numbers, and that f(t, y)f(t, y) is a function of tt and yy. Let t_n, ynt n ,y n satisfy the recursions t{n+1}=tn+h, \text{ and } \ y{n+1}= y_n+h f(t_n, y_n).t n+1 =t n +h, and y n+1 =y n +hf(t n ,y n ). Let RR be the rectangle in the tyty-plane given by: t_0\leq t\leq t_0+ at 0 ≤t≤t 0 +a, y_0- b\leq y\leq y_0+ by 0 −b≤y≤y 0 +b, and assume that |f(t, y)|\leq M∣f(t, y)∣≤M for all (t, y)(t, y) in RR. Finally, let \alpha = \min(a, b/ M)α=min(a, b/M). a) Prove that |y_j-y_0| \leq jhM∣y j −y 0 ∣≤jhM, as long as jh \leq \alphajh≤α. b) Conclude from (a) that the points (t_j, y_j)(t j ,y j ) all lie in RR as long as j \leq \alpha/h. j≤α/h

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

vertical=18cm horizontal=18cm hypotenuse=25.5

step-by-step explanation:

he answer is x=1x3 because x=1 to x=3

step-by-step explanation:

i'm not sure how to show the work for this

answer : c and d or 4(29 + 6)

4(29) + 4(6)