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3 years ago

1c - 20 = 4 - 3c pls help fast in school rn

Mathematics

2 answers:

nekit [7.7K]3 years ago

4 0

Answer:

4c-20= 4

+20 +20

4c= 24

/4 /4

c= 6

givi [52]3 years ago

4 0

Answer:

c = 1 btw can i be brainiest...

Step-by-step explanation:

First, rewrite the left side of the equation and group and combine like terms:

c + (4 − 3c) − 2 = 0

c + 4 − 3c − 2 = 0

−3c + c + 4 − 2 = 0

−3c + 1c + 4 − 2 = 0

(−3 + 1) c + (4 - 2) = 0

−2c + 2 = 0

Next, subtract

from each side of the equation to isolate the

term while keeping the equation balanced:

−2c + 2 − 2 = 0

−2 − 2c + 0 = −2

−2c = −2

Now, divide each side of the equation by

−2

to solve for

while keeping the equation balanced:

−2c - 2 = −2 − 2

−2c − 2 = 1

c = 1

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zavuch27 [327]

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3 years ago

The points plotted below are on the graph of a polynomial. Which of the following x-values best approximate roots of the polynom

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Answer:

The correct options are:

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8 0

3 years ago

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Goryan [66]

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8 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