It would be 67 because its the same decrease as the other.
Average rate of change over interval [a,b]: r=[f(b)-f(a)]/(b-a)
In this case the interval is [0,2], then a=0, b=2
r=[f(2)-f(0)]/(2-0)
r=[f(2)-f(0)]/2
1) First function: h(x)
r=[h(2)-h(0)]/2
x=2→h(2)=(2)^2+2(2)-6
h(2)=4+4-6
h(2)=2
x=0→h(0)=(0)^2+2(0)-6
h(0)=0+0-6
h(0)=-6
r=[h(2)-h(0)]/2
r=[2-(-6)]/2
r=(2+6)/2
r=(8)/2
r=4
2) Second function: f(x)
A function, f, has an
x-intercept at (2,0)→x=2, f(2)=0
and a y-intercept at (0,-10)→x=0, f(0)=-10
r=[f(2)-f(0)]/2
r=[0-(-10)]/2
r=(0+10)/2
r=(10)/2
r=5
3) Third function: g(x)
r=[g(2)-g(0)]/2
From the graph:
g(2)=6
g(0)=2
r=(6-2)/2
r=(4)/2
r=2
4) Fourth function: j(x)
r=[j(2)-j(0)]/2
From the table:
x=2→j(2)=-8
x=0→j(0)=4
r=(-8-4)/2
r=(-12)/2
r=-6
Answer:
Pairs
1) h(x) 4
2) f(x) 5
3) g(x) 2
4) j(x) -6
Hello,
f(n+1)=f(n)-2
f(1)=10
f(2)=8
f(3)=6
...
Answer:
<em>The second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
Step-by-step explanation:
We can't confirm the length of these diagonals based on the appearance of the figure, so let us apply Pythagorean Theorem;
This diagonal divides each figure ( square + rectangle ) into two congruent, right angle triangles ⇒ from which we may apply Pythagorean Theorem, where the diagonal acts as the hypotenuse;
5^2 + 5^2 = x^2 ⇒ x is the length of the diagonal,
25 + 25 = x^2,
x^2 = 50,
x = √50
Now the same procedure can be applied to this other quadrilateral;
3^2 + 7^2 = x^2 ⇒ x is the length of the diagonal,
9 + 49 = x^2,
x^2 = 58,
x = √58
<em>Therefore the second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
Answer:
i think but I am not sure so maybe ask somebody else but I think it is A.
Step-by-step explanation:
I just think it is
