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3 years ago

10(1+3x)=-20 and -5x-8(1+7x)=-8 With steps

Mathematics

1 answer:

SOVA2 [1]3 years ago

6 0

(1)

10(1+3x)=-20

Cross multiply.

10+30x=-20

Isolate x on one side. So you would subtract 10 on each side. and one side will cross each other out. leaving you with,

30x=-20-10

Subtract 10 from -20.

30x=-30

Divide each side by 30 to get x.

30x÷30=-30÷30

Therefore,

x=-1

(2)

-5x-8(1+7x)=-8

Cross multiply -8(1+7x)

-5x-8-56x=-8

Isolate all x's on one side. So you would add 8 on each side. and one side will cross each other out. leaving you with,

-5x-56x=-8+8

Add 8 to -8.

-5x-56x=0

Subtract -56x from -5x

-61x=0

Divide each side by -61

-61x÷-61=0÷-61

Therefore,

x=0

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Answer: 67

Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

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Answer questions a) and b)

Mariulka [41]

Answer:

$x\hookrightarrow -1\hookrightarrow 0\hookrightarrow 1\hookrightarrow 2\hookrightarrow 3\hookrightarrow 4\hookrightarrow 5$

$y\hookrightarrow 5\hookrightarrow 0\hookrightarrow -3\hookrightarrow -4\hookrightarrow -3\hookrightarrow 0\hookrightarrow 5$

The curve (B) matches the graph of y=x²-4x.

$----------\\Hope \ it\;helps$

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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>

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2 years ago