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Ratling [72]
3 years ago
6

Help plz Solve for f. 4(f + 2) = 16

Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

2

Step-by-step explanation:

4(f+2)=16

4f+8=16

4f=8

f=8/4

f=2

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-10=-15+5x heeeeeeeeeeeeeeeeeeeeeeelp me
Allisa [31]
The answer is -10=-15+5x x = 1
6 0
3 years ago
3.18x+2.6+0.9x+x=28<br> Solve please, brainliest will be awarded
Elis [28]

Answer:

x=5

Step-by-step explanation

3.18x+2.6+0.9x+x=28

3.18x+0.9x+x+2.6=28

5.08x+2.6=28

5.08x+2.6-2.6=28-2.6

5.08x=25.4

x=5

hope it helps

8 0
3 years ago
Read 2 more answers
Kenzie has. 4 pages of homework to do. If she can finish 1/3 of a page in one hour how many hours will her homework take?
Maru [420]

Answer:

It will take her 4 1/3 hours

Step-by-step explanation: so you know you needed to  add 4 and 1/3 and you turn the 4 into 4/1 and 1/3 and since it adding you have to have the same denominator and you multiply 1x3 and if you do it on the bottom you have to do it on top so you also do 4x3 and 4x3 is 12 and one 1x3 is 3 and so you don't have to have to do it on the other fraction because now they have the same denominator now and now you add 12/3 + 1/3 and you shall get 13/3 and then you divide 13 divided by 3 and you shall get 4 1/3 and that shall be your answer.

hope it helps you!

5 0
3 years ago
Triangle ABC has these side measurements: AB = 17 BC = 18 AC = 21 Order the angles of the triangle from largest measure to small
vodka [1.7K]

The angles are

  • AC
  • BC
  • AB

Hope this helps :)

6 0
2 years ago
Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)
Yakvenalex [24]

Answer:

\sin d = \frac{4}{7} ; \sin e = \frac{\sqrt{33} }{7}

\cos d = \frac{\sqrt{33} }{7} ; \cos e = \frac{4}{7}

\tan d = \frac{4}{\sqrt{33} } ; \tan e = \frac{\sqrt{33} }{4}

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

  • \sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \cos \alpha  = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \tan \alpha  = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }

__________________________________________________

According to the question ,

1) When α (alpha) = d

  • \sin d = \frac{4}{7}
  • \cos d = \frac{\sqrt{33} }{7}
  • \tan d = \frac{4}{\sqrt{33} }

2) When α (alpha) = e

  • \sin e = \frac{\sqrt{33} }{7}
  • \cos e = \frac{4}{7}
  • \tan e = \frac{\sqrt{33} }{4}

3 0
3 years ago
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