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telo118 [61]
3 years ago
12

Somebody help a girl out it’s my last question I need for my ixl. Please

Mathematics
1 answer:
zaharov [31]3 years ago
6 0

slope, rate of change is.

use this formula, to find right of change we can.

\frac{y.2 - y.1}{x.2 - x.1} = m

input two pairs of coordinates, we must.

use 0, 1 and 5, 4, we will.

4 - 1 / 5 - 0

3 / 5

3/5, the slope is.

hmm.

convert 3/5 to decimal, we must.

multiply both sides by 2, we can.

3 * 2 = 6

5 * 2 = 10

6/10, our new fraction is.

convert to decimal, we must.

6/10 = 0.6

0.6, our slope is.

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Please help this is on Prepworks
Olenka [21]

Answer:

D

Step-by-step explanation:

I hope it's the correct one

8 0
3 years ago
Can someone walk me through this?
Diano4ka-milaya [45]
So you can set up proportion.

\dfrac{7.5\ \text m}{76^\circ} = \dfrac C{360^\circ}

Here, you need to solve for C.

You can do that by multiple both sides by 360 and then you should get 360^\circ\cdot\dfrac{7.5\ \text m}{76^\circ} = C \approx \boxed{35.526\ \text m}

Hope this helps.
3 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
Please help me lol aaaaa
miss Akunina [59]

Answer:

−0.125 or - 1/8

Step-by-step explanation:

8 0
3 years ago
Please help 9/5a+12=5
vova2212 [387]
A = -35/9 is the answer
8 0
3 years ago
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