Answer:
see below for the tables
Step-by-step explanation:
The differential equation is separable, so the solution is ...

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The values for yn are y+y'·h = y+2xyh. We take the "absolute error" to be the (signed) difference between the calculated yn and the actual value y(x).
18a is correct, 18b is incorrect, 18c is correct, 18d is false, 18e is false, 18f is true, 19 is incorrect.
Answer:
Approximate length of the diagonal fencing needed is 45.25 cm.
Step-by-step explanation:
Here, each side of the square shaped land = 32 ft
Now, the diagonal fencing is made in side the plot so that the land is divided into two right triangles.
<u>In the triangle:</u>
Base side of the triangle = 32 cm
Perpendicular length of the triangle = 32 cm
Let us assume the hypotenuse of the triangle = k cm
⇒The length of the diagonal in the square = k cm
by <u>PYTHAGORAS THEOREM</u> in a right angled triangle:

⇒
or, k = 45.25 cm
Hence,the approximate length of the diagonal fencing needed to fence the square land is 45.25 cm.
1 ft = 12 inches......1 yd = 36 inches
carol has 54 inches
Tino has 7.5 ft......7.5 * 12 = 90 inches
baxter has 3.5 yds......3.5 * 36 = 126 inches
for a total of : (54 + 90 + 126) = 270 inches total <===
1 ft = 12 inches....so 270 inches = 270/12 = 22.5 ft <==
1 yd = 36 inches..so 270 inches = 270/36 = 7.5 yds <==
Dommain is input
range is output from given input (domain)
1,2
just input 1 for x and evaluate
then input 2 for x and evaluate
f(x)=3x
f(1)=3(1)=3
f(2)=3(2)=6
range=(3,6)
f(x)=x^2-2x-5
f(1)=1^2-2(1)-5=1-2-5=-6
f(2)=2^2-2(2)-5=-5
range=(-6,-5)
f(x)=(x+5)x^2
f(1)=(1+5)(1^2)=(6)(1)=6
f(2)=(2+5)(2^2)=(7)(4)=28
range=(6,28)
f(x)=4-x
f(1)=4-1=3
f(2)=4-2=2
range=(3,2)