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madreJ [45]
2 years ago
11

−1.2(3x − 3) = 4(3x +0.9) help pls

Mathematics
1 answer:
Viefleur [7K]2 years ago
6 0

Answer:

- 1.2(3x - 3) = 4(3x + 0.9) \\  - 3.6x + 3.6 = 12x + 3.6 \\  12x + 3.6x = 3.6 - 3.6 \\ 15.6x = 0 \\  \\ x =  \frac{0}{15.6}  \\  \\ x = 0

I hope I helped you^_^

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A credit card company charges 22% interest fee on any charges not paid at the end of the month.. If the bill totals $450 for thi
kifflom [539]
I think the answer is 450x .22= 99$
6 0
3 years ago
An item that cost $5.50 is marked up %60. What is the new price of the item after the markup? Show work please​
r-ruslan [8.4K]

Answer:

$8.80

Step-by-step explanation:

In order to find the new price, you need to add 60% of the original price (5.50) to the original price. You have two options:

1) multiply $5.50*0.6 to get 60% of the original price. You would get $3.30. Then add $3.30+$5.50, which equals $8.80. This way takes longer but shows you all the steps.

2) The shorter way is to simply multiply $5.50*1.6, which also equals $8.80. This works because the "1" automatically adds the 5.50 to 60% of the original price, this way you don't have to multiply then add.

If you don't understand how the second way works or your teacher taught you #1, just do it the first way.

8 0
3 years ago
Add. Write an integer or fraction in lowest terms.<br><br>7/8 +8/9​
8_murik_8 [283]

Answer:

127/72

Step-by-step explanation:

7/8+8/9

63/72+64/72

127/72

8 0
3 years ago
Read 2 more answers
Solve for x. Round your answer to the nearest tenth if necessary.
posledela

Answer: 15.9

Step-by-step explanation:

7 0
2 years ago
Look at this cylinder:
Sedbober [7]
  • Height=h=8cm
  • Radius=r=4cm

We know

\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2

Now

  • New Radius=2(4)=8cm
  • New Height=2(8)=16cm

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2

So

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}

\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}

Hence our correct option is Option C

6 0
2 years ago
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