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Virty [35]
2 years ago
11

I have attached the question below

Mathematics
1 answer:
aniked [119]2 years ago
4 0

\text{The perimeter of the semicircle} = \text{radius}(\pi+2)=3(\pi+2)~cm=(3\pi + 6) ~cm\\

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-8p^2+8p+7=9<br> how do i find the discriminant
m_a_m_a [10]
The discriminant can be found using the formula b^2-4ac 

First put your equation in standard form, where all your values are on one side, and just add f(x) or y in front of your equation. 
y= 8p^2-8p+2

The first value of your equation is a   (a=8)
The second term of your equation is b  (b=-8)
The last term of your equation is c  (c=2)

Plug in the values to the discriminant equation b^2-4ac 
5 0
3 years ago
3.) Find the quotient<br> 35)2,520<br> Answer:
AleksAgata [21]

Answer:

1.38888889

Step-by-step explanation:

Do the math on the calculator to get the answer.

3 0
3 years ago
How do convert fluid ounces into quarts
34kurt
<span>Quarts to fluid ounces volume units conversion factors are listed below. To find out how many ounces in quarts, multiply by the right factor or instead, use the converter below if converting between US fluid measurements.</span>
6 0
3 years ago
Read 2 more answers
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
The expression 3(x-9) is equivalent to<br> 3(x)+9.<br> 3(x)+3(9).<br> 3(x) – 9.<br> 3(x) – 3(9).
sukhopar [10]

Answer:

3(x) – 3(9)

Step-by-step explanation:

3(x-9)

Distribute the 3 to each term in the parentheses

3*x -3*9

3x -27

3 0
2 years ago
Read 2 more answers
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