Answer:
y=2x-8
or
f(x)=2x-8
Step-by-step explanation:
To write in slope-intercept form using this information, first use the point slope formula.
y-y1=m(x-x1)
m is slope.
y+4=2(x-2)
Use distributive property on the right side.
y+4=2x-4
Now, subtract 4 from both sides.
y+4-4=2x-4-4
y=2x-8
Written using function notation:
f(x)=2x-8
Hope this helps!
If not, I am sorry.
Answer:
A
Step-by-step explanation:
A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)
Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say
x.
The lateral faces are equilateral triangles of side length
x.
The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.
now we need to find
x. Consider the picture attached, showing one lateral face of the pyramid.
by the Pythagorean theorem:

thus,

thus:

(cm^2)
but

is exactly the base area, since the base is a square of sidelength =
x cm.
So, the total surface area = base area + lateral area =

cm^2
Answer:

cm^2
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
Answer:
d.100
Step-by-step explanation:
i just took the test