Answer:
(a) 
(b) 
(c) Test average for maths class after test 2 is greater
Step-by-step explanation:
Given
Solving (a): f(2)
We have:
Solving (b): g(2)
From the table:
when 
So:
Solving (c): Which is greater f(2) or g(2)
In (a) and (b),
Hence, test average for maths class is greater
Answer:
RTA= (x-2)·(x-11/2)/(x-2)(x-4)= (you can simplify again if you want by eliminating both (x-2)
(x-11/2)/(x-4)
Step-by-step explanation:
Ok we need to simplify the expression so:
x^2+3x-10= Bhaskara formula=
-3(±√9-4·1·(-10))/2·1=
X1=(-3+7/2)--> X1=2(R)---> (X-R)--->X-2
X2=(-3-7)/2 --> X2=11/2(R)---> (X-R)--->X-11/2
x^2-6x+8= Bhaskara formula=
6(±√36-4·1·8)/2·1=
X1=(6-2)/2=2--> X1=2(R)---> (X-R)--->X-2
X2=(6+2)/2=2--> X2=4(R)---> (X-R)--->X-4
so, The simplify expression is
(x-2)·(x-11/2)/(x-2)(x-4)=
Answer:
A
Step-by-step explanation:
<span>When we say about an object relative to something, this
only means that we are comparing an object’s characteristic to that of our
reference. So in this case, I think we are referencing Delia’s jump to someone
else. Say our reference has a long jump length of 5 meters, since Delia’s
relative long jump length is 0, this means that relative to the reference there
is no difference. Therefore, Delia’s jump length is also 5 meters.</span>
