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shtirl [24]
3 years ago
8

When you divide , we are

Mathematics
2 answers:
fomenos3 years ago
5 0
Could you be a little more specific.
strojnjashka [21]3 years ago
3 0
Could you restate the question in a different way
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Write the equation of the line in slope intercept form Slope=-4 Point=(2,-4)
Andrei [34K]

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.

5 0
2 years ago
Read 2 more answers
What is the equation of the graph below?
alex41 [277]

Answer:

A

Step-by-step explanation:

8 0
2 years ago
a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar
Margarita [4]
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: 

h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x

thus, 

Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2

thus:

x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\  \sqrt{3} }{3}=24 \sqrt{3}       (cm^2)

but x^{2} is exactly the base area, since the base is a square of sidelength = x cm.


So, the total surface area = base area + lateral area =  24 \sqrt{3}+72   cm^2


Answer: 24 \sqrt{3}+72   cm^2

4 0
3 years ago
Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (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,C
B(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,B
A (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,C
B(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>

8 0
3 years ago
During his 360-min. shift at the local grocery store, Jared spent 122 min. stocking shelves, 74 min. bagging groceries, and 62 m
sergeinik [125]

Answer:

d.100

Step-by-step explanation:

i just took the test

3 0
3 years ago
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